Question

One bag contains 2 white balls and 2 black balls and another contains 3 white and 5 black balls. A bag and one ball from that bag is chosen at random and ball is replaced. This trail is repeated five times. The probability that there are exactly four white balls drawn is approximately equal to $\frac{k}{10}$. Find the value of $k$.

Answer 1

First we will find the probability of drawing white ball in a single trial.
There is equal probability of each bag being chosen.
So the probability of white ball being chosen from either bag
$p=\frac{1}{2}.\frac{2}{4}+\frac{1}{2}.\frac{3}{8}=\frac{1}{4}+\frac{3}{16}=\frac{7}{16}.$
And the probability of black ball being chosen
$=1-\frac{7}{16}=\frac{9}{16}.$
Now the probability of 4 balls in 5 trials
${=}^5{C}_4.p^4.q$
$=5\times {\left(\frac{7}{16}\right)}^4\times \frac{9}{16}=\frac{108045}{1048576}=0\cdot 103.$
Therefore k=1.03 similar to k=1