Question

Bag 1 contains 3 red and 4 black balls and bag 2 contains 4 red and 5 black balls. Two balls are transferred at random from bag 1 to bag 2 then a ball is drawn from bag 2. The ball so drawn is found to be red in color. Find the probability that the transferred balls were both black.

Answer 1

Hint: Assume the events as \[{{E}_{1}},{{E}_{2}},{{E}_{3}}\] and A, where ball transferred is red, black, one red and one black and red ball from bag 2.Thus using Bayes theorem find the value of \[P\left( {}^{{{E}_{2}}}/{}_{A} \right)\].

Complete step-by-step answer:
It is said that bag 1 contains 3 red and 4 black balls.
Thus the total number of balls in bag 1 = 3 + 4 = 7 balls.
Similarly, in bag 2 there are 4 red and 5 black balls.
Thus the total number of balls in bag 2 = 4 + 5 = 9 balls.
It is said that 2 balls were transferred to bag 1and bag 2 in random.
Let us assume that \[{{E}_{1}},{{E}_{2}},{{E}_{3}}\] and A are such events. Such that,
\[{{E}_{1}}=\] Both transferred balls from bag 1 to bag 2 are red.
\[{{E}_{2}}=\] Both transferred balls from bag 1 to bag 2 are black.
\[{{E}_{3}}=\] out of the two transferred balls, one is red and the other is black.
A = drawing a read ball from bag 2.

What we require is \[P\left( {}^{{{E}_{2}}}/{}_{A} \right)\], i.e. probability that the transferred balls were
both black.
\[P\left( {{E}_{1}} \right)=\] probability that both the transferred balls are red
\[=\dfrac{^{3}{{C}_{2}}}{^{7}{{C}_{2}}}\].
This is of the form \[^{n}{{C}_{r}}=\dfrac{n!r!}{(n-r)!}\].
\[\therefore P\left( {{E}_{1}} \right)=\dfrac{^{3}{{C}_{2}}}{^{7}{{C}_{2}}}=\dfrac{{}^{3!}/{}_{(3-
2)!2!}}{{}^{7!}/{}_{(7-2)!2!}}=\dfrac{{}^{3!}/{}_{1!2!}}{{}^{7!}/{}_{5!2!}}=\dfrac{\dfrac{3\times
2!}{1!2!}}{\dfrac{7\times 6\times 5!}{5!\times 2}}=\dfrac{3\times 2}{7\times 6}=\dfrac{1}{7}\]
Thus we got \[P\left( {{E}_{1}} \right)={}^{1}/{}_{7}\].
Similarly \[P\left( {{E}_{2}} \right)=\] probability that the two balls are black from bag 1 to bag 2
\[P\left( {{E}_{2}} \right)=\dfrac{^{4}{{C}_{2}}}{^{7}{{C}_{2}}}=\dfrac{{}^{4!}/{}_{(4-2)!2!}}{7\times
3}=\dfrac{{}^{4!}/{}_{2!2!}}{7\times 3}=\dfrac{2\times 3}{7\times 3}=\dfrac{2}{7}\]
Value of \[P\left( {{E}_{2}} \right)={}^{2}/{}_{7}\].
Similarly, \[P\left( {{E}_{3}} \right)=\dfrac{^{3}{{C}_{1}}{{\times
}^{4}}{{C}_{1}}}{^{7}{{C}_{2}}}=\dfrac{3\times 4}{7\times 3}=\dfrac{4}{7}\]
Two balls are transferred from bag 1 to bag 2. Thus the total number of balls in bag 2 will be 11. A is
drawing of red ball from bag 2.
\[\therefore P\left( {}^{A}/{}_{{{E}_{1}}} \right)\] where two balls are added to bag 2 which are red. So
the total number of red balls in bag 2 will be 4 + 2 = 6.
\[\therefore P\left( {}^{A}/{}_{{{E}_{1}}} \right)={}^{6}/{}_{11}\].
Similarly, for \[P\left( {}^{A}/{}_{{{E}_{2}}} \right)\] = drawing of a red ball from bag 2.
\[\therefore P\left( {}^{A}/{}_{{{E}_{2}}} \right)={}^{4}/{}_{11}\].
\[P\left( {}^{A}/{}_{{{E}_{3}}} \right)\] one of the balls transferred is red, so the total number of red balls
in the bag 2 = 4 + 1 = 5.
\[\therefore P\left( {}^{A}/{}_{{{E}_{3}}} \right)={}^{5}/{}_{11}\].
We said that we need \[P\left( {}^{{{E}_{2}}}/{}_{A} \right)\]. We can solve it by using Bayes theorem.
Bayes theorem states that the relationship between the probability of hypothesis (H) before getting the evidence (E) is P(H)and the probability of the hypothesis after getting the evidence \[P\left( {H}/{E}\;
\right)\] is,
\[P\left( {H}/{E}\; \right)=\dfrac{P\left( {E}/{H}\; \right)}{P\left( E \right)}.P\left( H \right)\].

Similarly, \[P\left( {}^{{{E}_{2}}}/{}_{A} \right)=\dfrac{P\left( {{E}_{2}} \right).P\left( {}^{A}/{}_{{{E}_{2}}}
\right)}{P\left( {{E}_{1}} \right).\left( {}^{A}/{}_{{{E}_{1}}} \right)+P\left( {{E}_{2}} \right).\left(
{}^{A}/{}_{{{E}_{2}}} \right)+P\left( {{E}_{3}} \right).\left( {}^{A}/{}_{{{E}_{3}}} \right)}........(1)\]
\[\begin{align}
& P\left( {{E}_{1}} \right)={}^{1}/{}_{7},P\left( {{E}_{2}} \right)={}^{2}/{}_{7},P\left( {{E}_{3}}
\right)={}^{4}/{}_{7}. \\
& P\left( {}^{A}/{}_{{{E}_{1}}} \right)={}^{6}/{}_{11},P\left( {}^{A}/{}_{{{E}_{2}}}
\right)={}^{4}/{}_{11},P\left( {}^{A}/{}_{{{E}_{3}}} \right)={}^{5}/{}_{11}. \\
\end{align}\]
Then substitute all this value in equation (1).
\[\begin{align}
& P\left( {}^{{{E}_{2}}}/{}_{A} \right)=\dfrac{\dfrac{2}{7}\times \dfrac{4}{11}}{\left( \dfrac{1}{7}\times
\dfrac{6}{11} \right)+\left( \dfrac{2}{7}\times \dfrac{4}{11} \right)+\left( \dfrac{4}{7}\times \dfrac{5}{11}
\right)} \\
& =\dfrac{{}^{8}/{}_{77}}{{}^{6}/{}_{77}+{}^{8}/{}_{77}+{}^{20}/{}_{77}}=\dfrac{{}^{8}/{}_{77}}{{}^{\left(
6+8+20 \right)}/{}_{77}} \\
& =\dfrac{{}^{8}/{}_{77}}{{}^{34}/{}_{77}}=\dfrac{8}{77}\times
\dfrac{77}{34}=\dfrac{8}{34}=\dfrac{4}{17}. \\
\end{align}\]
Therefore, the probability that the transferred balls are both black = \[={}^{4}/{}_{17}\].
Thus we got the required probability.

Note: Bayes theorem provides a way to revise the existing predictions or theories given new or additional evidence. The Bayes theorem describes the probability of an event based on prior knowledge of conditions.