Name: Binomial Probability Distribution: Mean and Variance
Uploaded: 2019-07-19
Duration: 6 min 31 s
Description: A binomial distribution can be thought of as simply the probability of a SUCCESS or FAILURE outcome in an experiment or survey that is repeated multiple time...

Question

Bag $A$ contains $4$ white balls and $3$ black balls, while Bag $B$ contains $3$ white balls and $5$ black balls. Two balls are drawn from Bag $A$ and placed in Bag $B$ . Then, what is the probability of drawing a white ball from Bag $B$ ?

Accepted Answer

Let $E_{1,} E_{2}, E_{3}$ and $A$ be the events defined

$B$ Two white balls is transferred from bag $A$ to bag $B$

$E_{2} =$ Two black balls is transferred from bag $A$ to bag $B$

$E_{3} =$ One white and one black balls is transferred from bag $A$ to bag $B$

$A =$ One ball drawn from the bag is white.

$P (E_{1}) = \frac{4_{C_{2}}}{7_{C_{2}}} = \frac{4 \times 3}{\frac{2 \times 1}{\frac{7 \times 6}{2 \times 1}}} = \frac{2}{7}$ [Using ${}^{n}C_{r} = \frac{n!}{r! (n - r)!}$ ]

$P (E_{2}) = \frac{3_{C_{2}}}{7_{C_{2}}} = \frac{3 \times 2}{\frac{2 \times 1}{\frac{7 \times 6}{2 \times 1}}} = \frac{1}{7}$

$P (E_{3}) = \frac{4_{C_{1}} \times 3_{C_{2}}}{7_{C_{2}}} = \frac{4 \times 3}{7 \times 6} = \frac{4}{7}$

When $E_{1}$ has occurred, two white balls has already been transferred from bag $A$ to bag $B$ contains $5$ white balls and $5$ black balls.

$P (\frac{A}{E_{1}}) = \frac{5}{10}$

When $E_{2}$ has occurred two black balls has already been transferred from bag $A$ to bag $B$ contains $3$ white balls and $7$ black balls.

$P (\frac{A}{E_{2}}) = \frac{3}{10}$

When $E_{3}$ has occurred one black ball and one white balls has already been transferred from bag $A$ to bag $B$ contain $4$ white balls and $6$ black balls.

$P (\frac{A}{E_{3}}) = \frac{4}{10}$

By using law of total probability,

$P (A) = P (E_{1}) P (\frac{A}{E_{1}}) + P (E_{2}) P (\frac{A}{E_{2}}) + P (E_{3}) P (\frac{A}{E_{3}})$

$= \frac{2}{7} (\frac{5}{10}) + \frac{1}{7} (\frac{3}{10}) + \frac{4}{7} (\frac{4}{10})$

$= \frac{10}{70} + \frac{3}{70} + \frac{16}{70} = \frac{29}{70}$

Hence, the required probability is $\frac{29}{70}$ .

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