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

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

`$= (1)/(3) * ^(10)C_4 * 0.4^4 * 0.6^6$`

= 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)

`$=(1)/(3)* ^(10)C_4 * 0.4^4 * 0.6^6 + (1)/(3)* ^(10)C_4 * 0.5^4 * 0.5^6 + (1)/(3)* ^(10)C_4 * 0.2^4 * 0.8^6$`

`$=0.0836+0.0684+0.0294$`

`$=0.1813$`

P(A|B) =
`$(P(A \cap B))/(P(B))$`

`$=(0.0836)/(0.1813)$`

`$=0.4611$`