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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?
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Solution
Use value E1 = 5 or 6 to get
E2 = getting 1, 2, 3, or 4
n(E1) = 2, n(E2) = 4 n(s) = 6
P(E1) = `2/6 = 1/3`
P(E2) = `4/6 = 2/3`
Let E = one head is obtained
P(heads are obtained when the coin is tossed 3 times) = P(E|E1)
= P(HTT, THT, TTH)
= `3/8`
P(heads appear when the coin is tossed 1 time) = P(E|E2)
= `1/2`
Required process = `(1/2 xx 2/3)/((1/2 xx 2/3) + (3/8 xx 1/3))`
= `8/8 + 3`
= `8/11`
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