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A coin is tossed twice if the coin shows head it is tossed again but if it shows a tail then a die is tossed. If 8 possible outcomes are equally likely. Find the probability that the die shows a number greater than 4, if it is known that the first throw of the coin results in a tail.
2/3
1/4
2/5
1/3
S = { HH, HT, T1, T2, T 3, T 4, T 5, T6 } Let A be the event that the die shows a number greater than 4 and B be the event that the first throw of the coin results in a tail then, A = { T5, T6 } B = { T1, T2, T 3, T 4, T 5, T6 } Then P=P(A and B)/P(B)=(2/8)/(6/8)=1/3.
Hence, option 4 is the correct answer.
By: Amit Kumar ProfileResourcesReport error
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