Question

Suppose C1;C2;C3 are three didifferent biased coins, whose probability of heads equals 0.4, 0.5, and 0.2 respectively. Suppose coins are placed together in a box and you randomly picked a coin from the box. Flip the coin 10 times. Let A denote the event you randomly chose coin C1. Let B denote the event that you got exactly 4 heads out of the 10 coin flips.
Compute the following probabilities:

P(A∩B)
P(B)
P(A|B)

Answer 1

Solution :

$P(A \cap B) = P(\text{coin}\ C_1 \text{ chosen and 4 heads in 10 trails})$

                $= \frac{1}{3} \times ^{10}C_4 \times 0.4^4 \times 0.6^6$

               = 0.0836

P(B)= P(coin $C_1$ chosen and 4 heads in 10 trails) + P(coin $C_2$ chosen and 4 heads in 10 trail) + P(coin $C_3$ chosen and 4 heads in 10 trail)

     $=\frac{1}{3}\times ^{10}C_4 \times 0.4^4 \times 0.6^6 + \frac{1}{3}\times ^{10}C_4 \times 0.5^4 \times 0.5^6 + \frac{1}{3}\times ^{10}C_4 \times 0.2^4 \times 0.8^6$

    $=0.0836+0.0684+0.0294$

    $=0.1813$

P(A|B) = $\frac{P(A \cap B)}{P(B)}$

          $=\frac{0.0836}{0.1813}$

         $=0.4611$