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12 Jan 2016, 09:18
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C
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Difficulty:
45%
(medium)
Question Stats:
59% (01:19) correct
41%
(01:24)
wrong
based on 176
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In a game, a person is declared winner, if he gets 4 consecutive tails in five throws of a coin. What is the probability of A to win if he has got a tail in the first throw?
A) 1/5
B) 1/4
C) 1/8
D) 1/16
E) 1/6
A) 1/5
B) 1/4
C) 1/8
D) 1/16
E) 1/6
12 Jan 2016, 09:36
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First throw = Tails
For a person to win, the next 3 throws must be tails and the 5th throw can be heads or tails. i.e. TTTTH or TTTTT
So the person can win in 2 ways if the first throw is Tails.
Total number of outcomes for the next 4 throws = 2^4 = 16
Probability = 2/16 = 1/8
Answer: C
For a person to win, the next 3 throws must be tails and the 5th throw can be heads or tails. i.e. TTTTH or TTTTT
So the person can win in 2 ways if the first throw is Tails.
Total number of outcomes for the next 4 throws = 2^4 = 16
Probability = 2/16 = 1/8
Answer: C
12 Jan 2016, 11:09
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The question can also be tackled by looking at the probability of each throw.
The probability of each throw landing on tails is 1/2. He got one tail on the first throw, so he needs 3 more tails to win.
\(\frac{1}{2}*\frac{1}{2}*\frac{1}{2} = \frac{1}{8}\).
The 5th throw is irrelevant since it can be heads or tails and will not impact the outcome of the game.
Answer C
The probability of each throw landing on tails is 1/2. He got one tail on the first throw, so he needs 3 more tails to win.
\(\frac{1}{2}*\frac{1}{2}*\frac{1}{2} = \frac{1}{8}\).
The 5th throw is irrelevant since it can be heads or tails and will not impact the outcome of the game.
Answer C
