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Bunuel
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Bunuel
John tossed a fair coin 3 times. What is the probability that the outcome was “tails” exactly twice?
There are three possible scenarios here:

1) "Tails" on toss 1 and 2, and "Heads" on toss 3:
OR (1/2)*(1/2)*(1/2)=1/8

2) "Tails" on toss 1 and 3, and "Heads" on toss 2:
OR (1/2)*(1/2)*(1/2)=1/8

3) "Tails" on toss 3 and 2, and "Heads" on toss 1:
OR (1/2)*(1/2)*(1/2)=1/8

That's a total probability of (1/8)+(1/8)+(1/8)=3/8

Alternately, you can count: Only possible situation are: HTT, THT, TTH of the possible 8 outcomes. Hence 3/8.

Hope it helps :)
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John tossed a fair coin 3 times. What is the probability that the outcome was “tails” exactly twice?

A. 1/8
B. 1/4
C. 3/8
D. 1/2
E. 9/10

We need to determine the following:

P(TTH) = (1/2)^3 = 1/8

Since TTH can be arranged in 3!/2! = 3 ways, the probability is 3 x (1/8) = 3/8.

Answer: C
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Given that John tossed a fair coin 3 times and we need to find What is the probability that the outcome was “tails” exactly twice?

Coin is tossed 3 times => Total number of cases = \(2^3\) = 8

To find the cases in which the coin lands on tails exactly twice we need to select two places out of three _ _ _ in which we will get tails.

This can be done in 3C2 ways = \(\frac{3!}{2!*(3-2)!}\) = \(\frac{3*2!}{2!*1!}\) = 3 ways

=> P(2T) = \(\frac{3}{8}\)

So, Answer will be C
Hope it helps!

Watch the following video to learn How to Solve Probability with Coin Toss Problems

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