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Question

Suppose, a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1,2,3 or 4 she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1,2,3 or 4 with the die?


Solution

Let E1 the event that 5 or 6 is shown on die and E2 the event that 1,2,3, or 4 is shown on die. 

Then E1 and E2 are mutually exclusive and exhaustive events. and nE1=2, nE2 =4

Also, n(S)= 6 

PEE1 = P (exactly one head show up when coin is tossed thrice)

=P(HTT,THT,TTH)= 38 total number of events  =23=8)

PEE2  = P (head shows up when coin is tossed once) = 12

The probability that the girl threw, 1,2,3 or 4 with the die, if she obtained exactly one head, is given by PE2E

By using Baye's theorem, we obtain

P(E2E)=P(EE2)P(E2)P(EE1)P(E1)+P(EE2)P(E2)=12×2312×23+38×13=1313+18=88+3=811.

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