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Question 65 of 73

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

West Bengal WbchseWest Bengal HS (WBCHSE) Board 2025Subjective· 4mImportance★★★★★
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This is a Bayes'-theorem problem: work out P(one tail)P(\text{one tail}) under each die-outcome branch, then invert.

Let E1E_1 = "die shows 1 or 2" (probability 26=13\tfrac26=\tfrac13), and E2E_2 = "die shows 3, 4, 5, or 6" (probability 46=23\tfrac46=\tfrac23).

Under E1E_1: the coin is tossed 3 times; probability of exactly one tail is

P(1 tail∣E1)=(31)(12)3=38P(\text{1 tail}|E_1) = \binom{3}{1}\left(\frac12\right)^3 = \frac38

Under E2E_2: the coin is tossed once; probability of exactly one tail (i.e. a single tail) is

P(1 tail∣E2)=12P(\text{1 tail}|E_2) = \frac12

Apply Bayes' theorem to find P(E2∣1 tail)P(E_2\mid\text{1 tail}):

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