Question

A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag what is the probability that the drawn ball is:
A) Red
B) Black

Answer 1

Hint: According to the question we have to find the probability that a ball is at random from the bag which can be a red ball or a black ball. As given in the question the number of red balls in the bag is 3 and the number of the black balls in the bag is 5. So for this case first of all we would have to choose the required ball according to given in the question. If the red ball is drawn at random from the bag so number of ways to choose one red ball from the three red balls can be obtained with the help of the formula given below:
$c_r^n = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$…………………………(1)
Where n is the required number of balls and r is the total number of balls.
This will be our required event and our total outcomes are the sum of all balls contained in the bag with the help of the formula (1). Now we can find the required probability with the help of the formula given below.
Required probability $n(P) = \dfrac{{n(E)}}{{n(S)}}$………………………….(2)
Where, $n(E)$is the required event which is the number of red balls for the case given in the question and $n(S)$ is the total outcomes which is the sum of all the balls in the bag.
Now, as given in the question we find the probability of a black ball at random from the bag.

Complete step-by-step answer:
Given,
The number of red balls in the bag = 3
The number of black balls in the bag = 5
Step 1: As given in the question there are 3 red balls and if a red ball is drawn at random from the bag then the number of ways to choose one red ball from the three red balls can be obtained with the help of the formula (1) as mentioned in the solution hint.
Number of event or number of ways to choose one red ball = $\dfrac{{3!}}{{1!\left( {3 - 1} \right)!}}$
Step 2: On solving the obtained number of ways to choose one red ball.
$ = \dfrac{{3!}}{{2!}}$
$ = \dfrac{{3 \times 2!}}{{2!}}$
Hence, Number of event or number of ways to choose one red ball n(E) = 3
Step 3: Now will find the total outcomes means we have to choose the 1 red ball from the total numbers of balls contained in the bag.
Total number of balls in the bag = 4 red balls + 5 black balls
Therefore, Number of ways to choose 1 red ball from the total number of balls using the formula (1) as mentioned in the solution hint.
$
   = \dfrac{{8!}}{{1!\left( {8 - 1} \right)!}} \\
   = \dfrac{{8 \times 7!}}{{7!}} \\
   = 8 \\
 $
Number of total outcomes n(S) = 8
Step 4: Now, we can find the probability with the help of the formula (2) as mentioned in the solution hint.
Probability that a red ball is drawn from the bag,
$ = \dfrac{3}{8}$
Step 5: Now, according to the question we have to find the probability if a black ball is drawn at random from the bag. Hence the total number of ways to choose one black ball from the five black balls can be obtained with the help of the formula (1) as mentioned in the solution hint.
Number of event or number of ways to choose one black ball$ = \dfrac{{5!}}{{1!\left( {5 - 1} \right)!}}$
On solving, the obtained number of ways to choose one red ball.
$
   = \dfrac{{5!}}{{4!}} \\
   = \dfrac{{5 \times 4!}}{{4!}} \\
   = 5 \\
 $
Step 6: Now will find the total outcomes means we have to choose the 1 black ball from the total numbers of balls contained in the bag.
Therefore, Number of ways to choose 1 black ball from the total number of balls using the formula (1) as mentioned in the solution hint.
$
   = \dfrac{{8!}}{{1!\left( {8 - 1} \right)!}} \\
   = \dfrac{{8!}}{{7!}} \\
   = \dfrac{{8 \times 7!}}{{7!}} \\
   = 8 \\
 $
Step 7: Now, we can find the probability with the help of the formula (2) as mentioned in the solution hint.
Probability that a black ball is drawn from the bag,
$ = \dfrac{5}{8}$

Hence, the probability that a red ball drawn at random from the bag is $ = \dfrac{3}{8}$ and the probability that a black ball drawn at random from the bag is $ = \dfrac{5}{8}$

Note: Another method: As we that if the probability of an event to be occur is $p(E)$ then the probability on an event not to be occur will be $p(\overline E ) = 1 - p(E)$
Hence, if the probability that a red ball drawn at random from the bag is$p(E)$$ = \dfrac{3}{8}$
So, that the ball drawn from the bag not to be red is $
  p(\overline E ) = 1 - \dfrac{3}{8} \\
    \\
 $
Therefore,
Probability of the ball drawn from the bag not to be red is$p(\overline E ) = \dfrac{5}{8}$