Important Questions for CBSE Class 11 Maths Chapter 14 - Probability

1 Marks questions

1. Three coins are tossed simultaneously to list the sample space for the event.

Ans. S = [HHH, HHT, HTH, HTT, THH, THT, TTH, TTT]

2. Two dice are thrown simultaneously. Find the prob. of getting doublet.

Ans. ${\text{n  =  (s)  =  36}}$ ( be the sample space )

Let ‘E’ be the event of getting doublet

P(E) = $\dfrac{6}{{36}}$ (since E = ((1,1),(2,2),(3,3),(4,4),(5,5),(6,6))) = $\dfrac{1}{6}$

3. 20 cards are numbered from 1 to 20. One card is then drawn at random. What is the prob. of a prime number.

Ans. Let ‘E’ be the event of getting prime number and S be the sample space.

Therefore, ${\text{n(S)}}$= {1,2,3…..20}

${\text{n(E)}}$= {2,3,5,7,11,13,17,19}

Probability ${\text{P(E)  =  }}\dfrac{{{\text{n(E)}}}}{{{\text{n(S)}}}}{\text{  =  }}\dfrac{{\text{8}}}{{{\text{20}}}}{\text{  =  }}\dfrac{{\text{2}}}{{\text{5}}}$

4. If $\dfrac{3}{{10}}$ is the prob. that an event will happen, what is the prob. that it will not happen.

Ans. Let the event be ‘E’

Given, ${\text{P(E)}}$ = $\dfrac{3}{{10}}$

We know that, ${\text{P(E bar)}}$ = 1 - ${\text{P(E)}}$ = $1 - \dfrac{3}{{10}} = \dfrac{7}{{10}}$

5. If A and B are two mutually exclusive events such that P(A) = $\dfrac{1}{2}$, P(B) = $\dfrac{1}{3}$. Find P(A or B).

Ans. We know that,

       $P(AorB)~\text{ }=~\text{ }P(A)~\text{ }+~\text{ }P(B)~\text{ }-~\text{ }P(A\cap B)~~~$

                     $=\frac{1}{2}+\frac{1}{3}-\varnothing ~~~~$

                               $=\frac{5}{6}~$

6. If E and F are events such that P(E) = $\dfrac{1}{4}$, P(F) = $\dfrac{1}{2}$, P(E and F) = $\dfrac{1}{8}$. Find P(not E and not F).

Ans. P(E’$\cap$F’) = P(E$\cup$F)’

= 1 – P(E$\cup$F)$[\because {\text{P(E}} \cup {\text{F)}} = \dfrac{1}{4} + \dfrac{1}{2} + \dfrac{1}{8} = \dfrac{5}{8}]$

= 1 -$\dfrac{5}{8}$$= \dfrac{3}{8}$

7. A letter is chosen at random from the word ‘ASSASSINATION’. Find the prob. that the letter is a consonant.

Ans. P(consonant) = $\dfrac{7}{{13}}$

8. There are 4 men are 6 women on the city council. If one council member is selected for a committee at random, how likely is it that it is a women?

Ans. P(women member is selected) = $\dfrac{6}{{10}} = \dfrac{3}{5}$

9. 4 cards are drawn from a well shuffled deck of 52 cards. What is the prob. Of obtaining 3 diamonds and one spade.

Ans. $\frac{{ }^{13} \mathrm{C}_{3} \times{ }^{13} \mathrm{C}_{1}}{{ }^{52} C_{4}}$ = $\dfrac{{286}}{{20825}}$ (since, one ace out of 13 and 3 spades out of 13)

10. Describe the sample space. A coin is tossed and a die is thrown.

Ans. {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}

11. We wish to choose one child of 2 boys and 3 girls. A coin is tossed, if it comes up heads, a boy is chosen, otherwise a girl is chosen. Describe the sample space.

Ans. $\{ {\text{HB1,HB2,TG1,TG2,TG3}}\}$

12. What is the chance that a leap year, selected at random, will contain 53 Sundays.

Ans. The total number of days in a leap year is 366 and there are 52 complete weeks and two days over. The 2 days may be (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday) or (Saturday, Sunday).

P(a leap year has 53 sunday) = $\dfrac{2}{7}$

13. If P(A) = 0.6, P(B) = 0.4 and P(A$\cap$B) = 0, then the events are?

Ans. Exclusive and exhaustive.

14. In general the prob. of an event lie between?

Ans. 0 and 1.

15. A and B are two mutually exclusive events of an experiment. If P(not A) = 0.65,

 P(A$\cup$B) = 0.65, P(B) = k,  find k.

Ans.

 ${\text{P(A}} \cup {\text{B)  =  P(A)  +  P(B)}}$ 

 ${\text{P(A}} \cup {\text{B)  =  1  -  P(notA)  +  P(B)}}$ 

 ${\text{0}}{\text{.65  =  1  -  0}}{\text{.65  +  k}}$ 

 ${\text{k  =  0}}{\text{.30}}$  

16. A box contains 1 white and 3 identical black balls. Two balls are drawn at random in succession without replacement. Write the sample space of the experiment.

Ans. S = {WB, BW, BB}.

17. Three coins are tossed once. Find the probability at most two heads.

Ans. S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

         E = {HHT, THH, HTH, HTT, THT, TTH, TTT}

         P(E) = $\dfrac{7}{8}$

18. One card is drawn from a pack of 52 cards, find the probability that the drawn card is either red or king.

Ans. P $= \dfrac{{26 + 2}}{{52}}$ 

          $= \dfrac{{28}}{{52}}$ 

          $= \dfrac{7}{{13}}$  

19. Five cards are drawn from a well shuffled pack of 52 cards. Find the probability that all the five cards are hearts.

Ans. $\frac{{}^{\text{13}}{{\text{C}}_{\text{5}}}}{{}^{\text{52}}{{\text{C}}_{\text{5}}}}=\frac{33}{66640}$

20. From a deck of 52 cards four cards are accidently dropped. Find the chance that the missing cards should be one from each type.

Ans. $\frac{^{\text{13}}{{\text{C}}_{\text{1}}}{{\times }^{\text{13}}}{{\text{C}}_{\text{1}}}{{\times }^{\text{13}}}{{\text{C}}_{\text{1}}}{{\times }^{\text{13}}}{{\text{C}}_{\text{1}}}}{{}^{\text{52}}{{\text{C}}_{\text{4}}}}=\frac{2197}{20825}$

21. In a random sampling three items are selected from a lot. Each item is tested and classified as defective(D) and non-defective(H). Write the sample space.

Ans. S = {DDD, DDN, DND, NDD, DNN, NDN, NND, NNN}

22. Let ${\text{S = \{ W1,W2,W3,W4,W5,W6\} }}$ be the sample space. Is the probability to outcome valid.

  ${\text{W1}}$  ${\text{W2}}$  ${\text{W3}}$  ${\text{W4}}$  ${\text{W5}}$  ${\text{W6}}$

  $\dfrac{1}{6}$     $\dfrac{1}{6}$    $\dfrac{1}{6}$     $\dfrac{1}{6}$    $\dfrac{1}{6}$     $\dfrac{1}{6}$

Ans.  Yes, $\left[ {\because \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} = 1} \right]$

23. The odds in favour of an event are 3:5, find the probability of occurrence of this event.

Ans. P = $\dfrac{3}{8}$

24. What the probability that an ordinary year will have 53 Sundays.

Ans. $\dfrac{1}{7}$

25. If odds against an event are 7:9, find the probability of non-occurrence of this event.

Ans. $1 - \dfrac{9}{{16}} = \dfrac{{16 - 9}}{{16}} = \dfrac{7}{{16}}$

 4 Marks questions

1. A coin is tossed three times consider the following event A: No head appears, B: Exactly one head appears and C: At least two heads appears do they form a set of mutually exclusive and exhaustive events.

Ans. S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

From the given data, A={TTT}, B={HTT, THT, TTH}, C={HHT, HTH, THH, HHH}

${\text{A}} \cup {\text{B}} \cup {\text{C = S}}$

Therefore, A, B and C are exhaustive events.

Also, $\text{A}\cap \text{B = }\varnothing \text{,A}\cap \text{C=}\varnothing \text{,C}\cap \text{C=}\varnothing \text{,}$disjoint i.e. they are mutually exclusive.

2. A and B are events such that P(A) = 0.42, P(B) = 0.48, and P(A and B) = 0.16. Determine (i)P(not A) (ii) P(not B) (iii) P(A or B)

Ans.

$P(\operatorname{not} A)=1-P(A)=1-0.42=0.58$

$P($ not $B)=1-P(B)=1-0.48=0.52$

$P($ AorB $)=P(A)+P(B)-P(A \cap B)$

$=0.42+0.48-0.16$

$=0.74$

3. Find the prob. that when a hand of 7 cards is drawn from a well shuffled deck of 52 cards, it contains (i) all king (ii) 3 kings (iii) atleast 3 kings 

Ans. P(all king) = $\frac{{}^{\text{4}}{{\text{C}}_{\text{4}}}{{\times }^{\text{48}}}{{\text{C}}_{\text{3}}}}{{}^{\text{52}}{{\text{C}}_{\text{7}}}}=\frac{1}{7735}$

P(3 king) = $\frac{{}^{\text{4}}{{\text{C}}_{\text{3}}}{{\times }^{\text{48}}}{{\text{C}}_{\text{4}}}}{{}^{\text{52}}{{\text{C}}_{\text{7}}}}=\frac{9}{1547}$

P(atleast 3 kings) = P(3 king) + P(4 king) = 

$\dfrac{9}{{1547}} + \dfrac{1}{{7735}} = \dfrac{{46}}{{7735}}$

4. From a group of 2 boys and 3 girls, two children are selected at random. Describe the sample space associated with 

(i) ${\text{E1}}$: Both the selected children are boys.

(ii) ${\text{E2}}$: At least one selected child is a boy.

(iii) ${\text{E3}}$: one boy and one girl is selected.

(iv) ${\text{E4}}$: both the selected children are girls.

Ans. 

 ${\text{S = \{ B1B2, B1G1, B1G2, B1G3, B2G1, B2G2, B2G3, G1G2, G1G3, G2G3\} }}$ 

 ${\text{E1 = \{ B1B2\} }}$ 

 ${\text{E2 = \{ B1B2, B1G1, B1G2, B1G3, B2G1, B2G2, B2G3\} }}$ 

 ${\text{E3 = \{ B1G1, B1G2, B1G3, B2G1, B2G2, B2G3\} }}$ 

 ${\text{E4 = \{ G1G2, G1G3, G2G3\} }}$  

5. A book contains 100 pages. A page is chosen at random. What is the chance that the sum of the digit on the page is equal to 9.

Ans. ${\text{E}}$= {9, 18, 27, 36, 45, 54, 63, 72, 81, 90}

S = 100

P(E) = $\dfrac{{10}}{{100}}$$= \dfrac{1}{{10}}$

6. A pack of 8 tickets numbered from 1 to 50 is shuffled and the two tickets are drawn find the prob.

(i) Both the ticket drawn bear prime number.

(ii) Neither of the tickets drawn bear prime number.

Ans.

The prime numbers between 1 to 50 are,

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47

(i) P(both ticket bearing prime number) = $\frac{{}^{\text{15}}{{\text{C}}_{\text{2}}}}{{}^{\text{50}}{{\text{C}}_{\text{2}}}}=\frac{3}{35}$

(ii) P(neither of the tickets bear prime number) = $\frac{{}^{\text{35}}{{\text{C}}_{\text{2}}}}{{}^{\text{50}}{{\text{C}}_{\text{2}}}}=\frac{17}{35}$

7. In a class XI of a school 40% of students study mathematics and 30% study biology. 10% of the class study both mathematics and biology. If a student is selected at random from the class, find the prob. that he will be studying mathematics or biology.

Ans.

${\text{P(M)  =  }}\dfrac{{{\text{40}}}}{{{\text{100}}}}{\text{,P(B)  =  }}\dfrac{{{\text{30}}}}{{{\text{100}}}}$ 

${\text{P(M}} \cap {\text{B)  =  }}\dfrac{{{\text{10}}}}{{{\text{100}}}}$ 

${\text{P(M}} \cup {\text{B)  =  P(M)  +  P(B)  -  P(M}} \cap {\text{B)}}$ 

${\text{ = }}\dfrac{{{\text{40}}}}{{{\text{100}}}}{\text{ + }}\dfrac{{{\text{30}}}}{{{\text{100}}}}{\text{ - }}\dfrac{{{\text{10}}}}{{{\text{100}}}}{\text{  =  0}}{\text{.6}}$  

8. A hockey match is played from 3pm to 5pm. A man arrives late for the match what is the prob. that he misses the only goal of the match which is scored at the ${{20}_{th}}$ minute of the match?

Ans.

Total time = ${\text{3pm  -  5pm  =  2hr  =  120min}}$

He can see the goal only if he arrives within initial 20 minutes.

P(he see the goal) = $\dfrac{{20}}{{120}} = \dfrac{1}{6}$

P(not see the goal) = $1 - \dfrac{1}{6} = \dfrac{5}{6}$

9. In a single throw of two dice, find that prob. that neither a doublet nor a total of 10 will appear.

Ans.

Let the sample space be ‘${\text{S}}$’ and ${\text{E1}}$ be event of getting doublet and E2 be the event of getting a total of 10.

${\text{E1  =  \{ (1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\} }}$ 

${\text{E2  =  \{ (4,6),(5,5),(6,4)\} }}$ 

${\text{n(S)  =  36}}$ 

${\text{P(E1)  = }}\dfrac{{\text{6}}}{{{\text{36}}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{6}}}$ 

${\text{P(E2)  = }}\dfrac{{\text{3}}}{{{\text{36}}}}{\text{ = }}\dfrac{{\text{1}}}{{{\text{12}}}}$ 

${\text{P(E1}} \cap {\text{E2)  =  1}}$ 

${\text{P(E1}} \cup {\text{E2)  =  }}\dfrac{{\text{2}}}{{\text{9}}}$ 

${\text{P(E1'}} \cap {\text{E2')  =  P(E}} \cup {\text{E2)'}}$ 

${\text{ =  1  -  P(E1}} \cup {\text{E2)}}$ 

${\text{ = 1 - }}\dfrac{{\text{2}}}{{\text{9}}}{\text{ = }}\dfrac{{\text{7}}}{{\text{9}}}$  

10. The prob that a person will get an electrification contract is $\dfrac{2}{5}$ and the prob. that he will not get a plumbing contract is $\dfrac{4}{7}$. If the prob. of getting at least one contract is$\dfrac{2}{3}$,  what is the prob. that he will get both?

Ans.

Let A = an event of getting electrification contract

B = an event of getting plumbing contract

${\text{P(A)  =  }}\dfrac{{\text{2}}}{{\text{5}}}{\text{, P(notB)  =  }}\dfrac{{\text{4}}}{{\text{7}}}{\text{,}}$ 

${\text{P(B)  =  1 - }}\dfrac{{\text{4}}}{{\text{7}}}{\text{  =  }}\dfrac{{\text{3}}}{{\text{7}}}$ 

${\text{P(A}} \cap {\text{B)  =  }}\dfrac{{\text{2}}}{{\text{3}}}$ 

${\text{P(A}} \cup {\text{B)  =  P(A)  +  P(B)  -  P(A}} \cap {\text{B)}}$ 

${\text{ = }}\dfrac{{\text{2}}}{{\text{5}}}{\text{ + }}\dfrac{{\text{3}}}{{\text{7}}}{\text{ - }}\dfrac{{\text{2}}}{{\text{3}}}{\text{ = }}\dfrac{{{\text{17}}}}{{{\text{105}}}}$  

11. In a town of 6000 people, 1200 are over 50 yr. old and 2000 are females. It is known that 30% of the females are over 50 yr. What is the prob. that a randomly chosen individual from the town is either female of over 30 yr.

Ans. Let A1 be the event that the person is a female and A2 be the event that the person is 50 yr. old.

${\text{n(A1) = 2000,n(A2) = 1200}}$ 

${\text{n(A1}} \cap {\text{A2) = }}$30% of 2000 = $\dfrac{{30}}{{100}} \times 2000 = 600$${\text{n(A1}} \cup {\text{A2)  =  n(A1)  +  n(A2)  -  n(A1}} \cap {\text{A2)}}$ 

${\text{ = 2000  +  1200  -  600  =  2600}}$ 

${\text{P(A1}} \cup {\text{A2) = }}\dfrac{{{\text{2600}}}}{{{\text{6000}}}}{\text{ = }}\dfrac{{{\text{13}}}}{{{\text{30}}}}$  

12. In a class of 60 students, 30 opted for NCC, 32 opted for NSS, 24 opted for both NCC and NSS. If one of these students is selected at random. Find the prob. that

(i) The student opted for NCC or NSS.

(ii) The student has opted neither NCC nor NSS.

(iii) The student has opted NSS but not NCC.

Ans. Let ‘A’ be the student opted for NCC and ‘B’ be the student opted for NSS 

${\text{P(A) = }}\dfrac{{{\text{30}}}}{{{\text{60}}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{, P(B) = }}\dfrac{{{\text{32}}}}{{{\text{60}}}}{\text{ = }}\dfrac{{\text{8}}}{{{\text{15}}}}$ 

${\text{P(A}} \cap {\text{B) = }}\dfrac{{{\text{24}}}}{{{\text{60}}}}{\text{ = }}\dfrac{{\text{2}}}{{\text{5}}}$ 

(i) We know that $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$

$\therefore \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{1}{2}+\frac{8}{15}-\frac{2}{5}=\frac{15+16-12}{30}=\frac{19}{30}$

Thus the probability that the selected student has opted for NCC or NSS is $\frac{19}{30}$

(ii)

${\text{P(A'}} \cap {\text{B') = P(A}} \cup {\text{B)'}}$ 

${\text{ = 1 - P(A}} \cup {\text{B)}}$ 

${\text{ = 1 - }}\dfrac{{{\text{19}}}}{{{\text{30}}}}{\text{ = }}\dfrac{{{\text{11}}}}{{{\text{30}}}}$  

(iii)

${\text{P(B - A)  =  P(B)  -  P(A}} \cap {\text{B)}}$ 

$= \dfrac{8}{{15}} - \dfrac{2}{5} = \dfrac{2}{{15}}$  

13. Two students Anil and Ashima appeared in an examination. The probability That Anil will qualify the examination is 0.05 and that Ashima will qualify the examination is 0.10. the probability that both will qualify the examination is 0.02. find the probability that

(a). Both Anil and Ashima will qualify the examination

(b). At least one of them will not qualify the examination and

(c). Only one of them will qualify the examination.

Ans. Let the events E and F denote that Anil and Ashima will pass in the examination respectively.${\text{P(E) = 0}}{\text{.05, P(F) = 0}}{\text{.10, P(E}} \cap {\text{F) = 0}}{\text{.02}}$

(a) 

${\text{P(E'}} \cap {\text{F')  =  P(E}} \cup {\text{F)'}}$ 

${\text{ = 1  -  P(E}} \cup {\text{F)}}$ 

${\text{ = 1 - }}\left[ {{\text{P(E)  +  P(F)  -  P(E}} \cap {\text{F)}}} \right]$ 

${\text{ = 1  -  0}}{\text{.13  =  0}}{\text{.87}}$  

(b) P(at least one of them will not qualify) = 1 – P(both of them will qualify)

=1 – 0.02 = 0.98

(c) P(only one of them will qualify) =

${\text{P(E}} \cap {\text{F')  +  P(E'}} \cap {\text{F)}}$ 

$  {\text{ = P(E)  -  P(E}} \cap {\text{F)  +  P(F)  -  P(E}} \cap {\text{F)}}$ 

${\text{ = 0}}{\text{.05  -  0}}{\text{.02  +  0}}{\text{.10  -  0}}{\text{.02}}$ 

${\text{ = 0}}{\text{.11}}$  

14. Out of 100 students, two sections of 40 and 60 are formed. If you and your friend are among the 100 students what is the probability that

(a) You both enter the same section

(b) You both enter the different section.

Ans. The 2 sections are formed in $^{{\text{100}}}{\text{C_40}}$or $^{{\text{100}}}{\text{C_60}}$ways.

(a) P(both enter the same section)

$=\text{ }\frac{{}^{\text{40}}{{\text{C}}_{\text{2}}}}{{}^{\text{100}}{{\text{C}}_{\text{2}}}}\text{+ }\frac{{}^{\text{60}}{{\text{C}}_{\text{2}}}}{{}^{\text{100}}{{\text{C}}_{\text{2}}}}\text{= }\frac{\text{17}}{\text{33}}$

(b) Required probability =

$\frac{{}^{\text{40}}{{\text{C}}_{\text{1}}}{{\times }^{\text{60}}}{{\text{C}}_{\text{1}}}}{{}^{\text{100}}{{\text{C}}_{\text{2}}}}\text{ = }\frac{\text{16}}{\text{33}}$

15. There are three mutually exclusive and exhaustive events ${\text{E1,}}\,{\text{E2}}\,$and ${\text{E3}}$. The odds are 8:3 against ${\text{E1}}$and 2:5 in favours of ${\text{E2}}$ find the odd against ${\text{E3}}$.

Ans.

Odds against E1  are 8:3. So, odds in favour of E1  are 3:8.

$\therefore {\text{P(E1) = }}\dfrac{{\text{3}}}{{{\text{3 + 8}}}}{\text{ = }}\dfrac{{\text{3}}}{{{\text{11}}}}{\text{,}}\,\,\,{\text{P(E2) = }}\dfrac{{\text{2}}}{{{\text{2 + 5}}}}{\text{ = }}\dfrac{{\text{2}}}{{\text{7}}}$ 

${\text{P(E1) }}\,{\text{ +  }}\,{\text{P(E2)}}\,{\text{ +  P(E3)  =  1}}\left[ {{\text{E1,E2}}\,{\text{and}}\,{\text{E3}}\,\,{\text{are}}\,{\text{mutually}}\,{\text{exclusive}}\,{\text{and exhaustive}}} \right]$ 

${\text{ = 1 - }}\dfrac{{\text{3}}}{{{\text{11}}}}{\text{ - }}\dfrac{{\text{2}}}{{\text{7}}}{\text{ = }}\dfrac{{{\text{34}}}}{{{\text{77}}}}$  

Odds against E3  are 

${\text{ = }}\dfrac{{{\text{1}}\,{\text{ - }}\,{\text{P(E3)}}}}{{{\text{P(E3)}}}}$ 

${\text{ = }}\dfrac{{{\text{1}}\,{\text{ - }}\,\dfrac{{{\text{34}}}}{{{\text{77}}}}}}{{{\text{77}}}}\,{\text{ = }}\,\dfrac{{{\text{43}}}}{{{\text{34}}}}$  

16 . If an entrance test that is graded on the basis of two examinations, the probability of a randomly chosen student passing the first examination is 0.8 and the probability of passing the second examination is 0.7. The probability of passing at least one of them is 0.95. What is the probability of passing both?

Ans. A: Student passes first examination

         B: Student passes second examination

${\text{P(A)}}\,{\text{ = }}\,{\text{0}}{\text{.8,}}\,{\text{P(B)}}\,{\text{ = }}\,{\text{0}}{\text{.7}}$ 

${\text{P(A}} \cup {\text{B)}}\,{\text{ = }}\,{\text{0}}{\text{.95}}$ 

${\text{P(A}} \cap {\text{B)}}\,{\text{ = }}\,{\text{?}}$ 

${\text{P(A}} \cup {\text{B)}}\,{\text{ = }}\,{\text{P(A)}}\,{\text{ + }}\,{\text{P(B)}}\,{\text{ - }}\,{\text{P(A}} \cap {\text{B)}}$ 

${\text{0}}{\text{.95}}\,\,{\text{ = }}\,{\text{0}}{\text{.8}}\,{\text{ + }}\,{\text{0}}{\text{.7}}\,{\text{ - }}\,{\text{P(A}} \cap {\text{B)}}$ 

${\text{0}}{\text{.55}}\,{\text{ = }}\,{\text{P(A}} \cap {\text{B)}}$  

17. One card is drawn from a well shuffled deck of 52 cards. If each outcome is equally likely calculate the probability that the card will be

(i) a diamond (ii) Not an ace (iii) A black card (iv) Not a diamond

Ans. (i) Required probability

$= \dfrac{{13}}{{52}} = \dfrac{1}{4}$

(ii) Required probability 

=$1\, - \,\dfrac{4}{{52}}\, = \,1\, - \,\dfrac{1}{{13}}\, = \,\dfrac{{12}}{{13}}$

(iii) Required probability

$= \dfrac{{26}}{{52}}\, = \,\dfrac{1}{2}$

(iv) Required probability

$= 1\, - \,\dfrac{1}{4}\, = \,\dfrac{3}{4}$

18. In a lottery, a person chooses six different natural no. at random from 1 to 20 and if these six no. match with six no. already fixed by the lottery committee, he wins the prize. What is the probability of winning the prize in the game? 

Ans. 6 natural numbers can be chosen from 20 in $^{20}C6$ ways out of these only one choice will match the six no. already by the committee

P(The person wins the prize) 

$=\text{ }\frac{\text{1}}{{}^{\text{20}}{{\text{C}}_{\text{6}}}}=\frac{1}{38760}$

19. From the employees of the company, 5 persons are elected to represent them in the managing committee of the company.

A person is selected at random from this group as a spoke person. What is the probability that a spoke person will be either male or over 35 yr.

Ans. Let the events be A: Spoke person is a male

B: Spoke person is over 35 yr.

${\text{P(A)}}\,{\text{ = }}\,\dfrac{{\text{3}}}{{\text{5}}}$ 

${\text{P(B)}}\,{\text{ = }}\,\dfrac{{\text{2}}}{{\text{5}}}$ 

${\text{P(A}} \cap {\text{B)}}\,{\text{ = }}\,\dfrac{{\text{1}}}{{\text{5}}}$ 

${\text{P(A}} \cup {\text{B)}}\,{\text{ = }}\,{\text{P(A)}}\,{\text{ + }}\,{\text{P(B)}}\,{\text{ - }}\,{\text{P(A}} \cap {\text{B)}}$ 

${\text{ = }}\dfrac{{\text{3}}}{{\text{5}}}{\text{ + }}\dfrac{{\text{2}}}{{\text{5}}}{\text{ - }}\dfrac{{\text{1}}}{{\text{5}}}\,{\text{ = }}\,\dfrac{{\text{4}}}{{\text{5}}}$  

20. A die has two faces each with no. 1three faces each with no. 2and one face with no. 3 if the die is rolled once, determine

(i) P(2) (ii) P(1 or 3) (iii) P(not 3)

Ans. Let A, B and C be the events

A: getting a face with no. 1

B: getting a face with no.2

C: getting a face with no.3

${\text{P(A)}}\,{\text{ = }}\,\dfrac{{\text{2}}}{{\text{6}}}\,{\text{ = }}\,\dfrac{{\text{1}}}{{\text{3}}}$ 

${\text{P(B)}}\,{\text{ = }}\,\dfrac{{\text{3}}}{{\text{6}}}\,{\text{ = }}\,\dfrac{{\text{1}}}{{\text{2}}}$ 

${\text{P(C)}}\,{\text{ = }}\,\dfrac{{\text{1}}}{{\text{6}}}$ 

${\text{(i)}}\,{\text{P(1}}\,{\text{or}}\,{\text{3)}}\,{\text{ = }}\,{\text{P(1)}}\,{\text{ + }}\,{\text{P(3)}}$ 

${\text{ = }}\dfrac{{\text{1}}}{{\text{3}}}\,{\text{ + }}\,\dfrac{{\text{1}}}{{\text{6}}}\,{\text{ = }}\,\dfrac{{\text{1}}}{{\text{2}}}$ 

${\text{(ii)}}\,{\text{P(not}}\,{\text{3)}}\,{\text{ = }}\,{\text{1}}\,{\text{ - }}\,\dfrac{{\text{1}}}{{\text{6}}}\,{\text{ = }}\,\dfrac{{\text{5}}}{{\text{6}}}$  

21. Find the probability that in a random arrangement of the letters of the word UNIVERSITY the two I’s come together.

Ans. From the word ‘UNIVERSITY’, the total number of words that can be formed is $\dfrac{{10!}}{{2!}}$.

Consider 2I’s as one letter, No. of ways of arranging in which 2I’s are together is

$= \dfrac{{9!}}{{\dfrac{{10!}}{{2!}}}} = \dfrac{1}{5}$

22. A bag contains 50 tickets no. 1, 2, 3,….., 50 of which 5 are drawn at random and arranges in ascending order of magnitude $(x1\, < \,x2\, < \,x3\, < \,x4\, < \,x5)$ find the probability that $x3$=30.$

$\text{Ans}\text{. Outof }50\text{ , }5\text{ tickets can be drawn in}$$^{\text{50}}{{\text{C}}_{\text{3}}}$ways

Since, ${\text{(x1}}\,{\text{ < }}\,{\text{x2}}\,{\text{ < }}\,{\text{x3}}\,{\text{ < }}\,{\text{x4}}\,{\text{ < }}\,{\text{x5)}}$

And ${\text{x3}}$ = 30

${\text{x1}}\,{\text{ < }}\,{\text{x2}}\,{\text{ < }}\,{\text{30}}$

i.e., x1 and x $\text{should be from tickets no}\text{. }1\text{to }29\text{ =}$$^{\text{29}}{{\text{C}}_{\text{2}}}\text{ways}\text{. }$$\text{Remaining }2\text{ should be from }20\text{ tickets i}\text{.e}\text{. from }31\text{ to }50\text{ }=$$^{\text{20}}{{\text{C}}_{\text{2}}}$$\text{ways Favorable case}=$$^{29}{{C}_{2}}{{\times }^{20}}{{C}_{2}}$$\text{Required probability}=$$\frac{{}^{\text{29}}{{\text{C}}_{\text{2}}}\times {{}^{\text{20}}}{{\text{C}}_{\text{2}}}}{{}^{\text{50}}{{\text{C}}_{\text{5}}}}\text{ = }\frac{\text{551}}{\text{15134}}$

6 Marks questions

1. Three letters are dictated to three persons and an envelope is addressed to each of them, those letters are inserted into the envelopes at random so that each envelope contains exactly one letter. Find the prob. that at least one letter is in its proper envelope.

Ans. Let the 3 letters be ${\text{A1,}}\,{\text{A2}}\,{\text{and}}\,{\text{A3}}$ and the three envelops be ${\text{E1,}}\,{\text{E2}}\,{\text{and}}\,{\text{E3}}$.

Ways of putting letters in 3 envelops is $^{\text{3}}{{\text{P}}_{3}}\text{ = }\text{6}$

Ways in which none of the letters is put in proper envelope = 2

Probability:

P(at least one letters in put into proper envelope) = 1 – P(none letters is put into proper envelope)

$= 1\, - \,\dfrac{2}{6}$ 

$= \dfrac{2}{3}$  

2. If a 4 digit number greater than 5,000 are randomly formed the digits 0,1,3,5 and 7 what is the prob. of forming a number divisible by 5 when

(i) The digits are repeated (ii) The repetition of digits is not allowed.

Ans. 

Thousand place can be filled in 2 ways and other places can be filled in 5 ways for digits greater than 5000

No. 40 can be formed = $2 \times 5 \times 5 \times 5\, = \,250$

If the number is divisible by 5, both unit and thousand place can be filled in 2 ways.

No. formed = $2 \times 5 \times 5 \times 2\, = \,100$

Probability = $\dfrac{{100}}{{250}}\, = \,\dfrac{2}{5}$

(ii) Digit not repeated

Thousand place can be filled in 2 ways

4 digit number greater than 5,000 = $2 \times 4 \times 3 \times 2\, = \,48$

Favorable case = $1 \times 3 \times 2 \times 2\, + \,1 \times 3 \times 2 \times 1$

7 and 5 at thousand place

=12 + 6

=18

Probability = $\dfrac{{18}}{{48}}\, = \,\dfrac{3}{8}$

3. 20 cards are numbered from 1 to 20. One card is drawn at random what is the prob. that the number on the card drawn is 

(i) A prime no. (ii) An odd no. (iii) A multiple of 5 (iv) Not divisible by 3

Ans. Let the sample space be ‘S’

S = {1, 2, 3, 4,……., 20}

Let the events ${\text{E1,}}\,{\text{E2,}}\,{\text{E3}}\,{\text{and}}\,{\text{E4}}$ be of getting a prime no., an odd no., multiple of 5 and not divisible by 3 respectively.

${\text{P(E1)}}\,{\text{ = }}\,\dfrac{{\text{8}}}{{{\text{20}}}}{\text{ = }}\dfrac{{\text{2}}}{{\text{5}}}{\text{,}}\,{\text{E1}}\,{\text{ = }}\,{\text{\{ 2,}}\,{\text{3,}}\,{\text{5,}}\,{\text{7,}}\,{\text{11,}}\,{\text{13,}}\,{\text{17,}}\,{\text{19\} }}$ 

${\text{P(E2)}}\,{\text{ = }}\,\dfrac{{{\text{10}}}}{{{\text{20}}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{,}}\,{\text{E2}}\,{\text{ = }}\,{\text{\{ 1,}}\,{\text{3,}}\,{\text{5,}}\,{\text{7,}}\,{\text{9,11,}}\,{\text{13,}}\,{\text{15,}}\,{\text{17,}}\,{\text{19\} }}$ 

${\text{P(E3)}}\,{\text{ = }}\,\dfrac{{\text{4}}}{{{\text{20}}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{5}}}{\text{,E3}}\,{\text{ = }}\,{\text{\{ 5,}}\,{\text{10,}}\,{\text{15,}}\,{\text{20\} }}$ 

${\text{P(E4)}}\,{\text{ = }}\,\dfrac{{{\text{14}}}}{{{\text{20}}}}{\text{ = }}\dfrac{{\text{7}}}{{{\text{10}}}}{\text{,}}\,{\text{E4}}\,{\text{ = }}\,{\text{\{ 1,}}\,{\text{2,}}\,{\text{4,}}\,{\text{5,}}\,{\text{7,}}\,{\text{8,}}\,{\text{10,}}\,{\text{11,}}\,{\text{13,}}\,{\text{14,}}\,{\text{16,}}\,{\text{17,}}\,{\text{19,}}\,{\text{20\} }}$  

4. In a single throw of three dice. Find the prob. of getting

(i) A total of 5 (ii) A total of at most 5

Ans. Let the sample space be ’S’ and the event of total of 5 be ‘E1’

(i)

${\text{E1}}\,{\text{ = }}\,{\text{\{ (1,}}\,{\text{1,}}\,{\text{3),(1,}}\,{\text{3,}}\,{\text{1),(3,}}\,{\text{1,}}\,{\text{1),(1,}}\,{\text{2,}}\,{\text{2),(2,}}\,{\text{1,}}\,{\text{2),(2,}}\,{\text{2,}}\,{\text{1)\} }}$ 

 $\text{S}\text{ = }\text{6}\times \text{6}\times \text{6}\text{ = }\text{216}$

 ${\text{P(E1)}}\,{\text{ = }}\,\dfrac{{{\text{n(E1)}}}}{{{\text{n(S)}}}}{\text{ = }}\dfrac{{\text{6}}}{{{\text{216}}}}{\text{ = }}\dfrac{{\text{1}}}{{{\text{26}}}}$  

(ii) 

 ${\text{E2}}\,{\text{ = }}\,{\text{\{ (1,}}\,{\text{1,}}\,{\text{1),(1,}}\,{\text{1,}}\,{\text{2),(1,}}\,{\text{2,}}\,{\text{1),(2,}}\,{\text{1,}}\,{\text{1),(1,}}\,{\text{1,}}\,{\text{3),(1,}}\,{\text{3,}}\,{\text{1),(3,}}\,{\text{1,}}\,{\text{1),(1,}}\,{\text{2,}}\,{\text{2),(2,}}\,{\text{1,}}\,{\text{2),(2,}}\,{\text{2,}}\,{\text{1)\} }}$ 

  ${\text{P(E2)}}\,{\text{ = }}\,\dfrac{{{\text{10}}}}{{{\text{216}}}}\,{\text{ = }}\,\dfrac{{\text{5}}}{{{\text{108}}}}$ 

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