Suppose C1;C2;C3 are three didifferent biased coins, whose probability of heads equals 0.4, 0.5, and 0.2...
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)
Homework Answers
1)
P(A n B) =P(Coin C1 chosen and 4 heads in 10
trails)=(1/3)*((10C4)*(0.4)4(0.6)6
=0.0836
2)
P(B) =P(Coin C1 chosen and 4 heads in 10 trails)+P(Coin C2 chosen and 4 heads in 10 trails)+P(Coin C3 chosen and 4 heads in 10 trails)
=(1/3)*((10C4)*(0.4)4(0.6)6 +(1/3)*((10C4)*(0.5)4(0.5)6 +(1/3)*((10C4)*(0.2)4(0.8)6
=0.0836+0.0684+0.0294
=0.1813
3)
therefore P(A|B) =P(A n B)/P(B) =0.0836/0.1813 =0.4611
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