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John tossed a fair coin 3 times. What is the probability that the outc

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John tossed a fair coin 3 times. What is the probability that the outc  [#permalink]

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New post 25 Sep 2016, 10:02
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A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

72% (00:59) correct 28% (00:57) wrong based on 194 sessions

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Re: John tossed a fair coin 3 times. What is the probability that the outc  [#permalink]

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New post 25 Sep 2016, 12:49
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Bunuel wrote:
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


Total number of outcomes = 2*2*2 = 8
Number of favorable outcomes = Number of ways two places (for tails) can be chosen out of 3 slots = 3C2 = 3
So probability = 3/8

Alternative
Probability of getting tail in single toss = 1/2
Probability of getting head in single toss = 1/2

Probability of getting First Tail = 1/2
Probability of getting Second tail (Such that first tail has occurred, this incidentally is also the probability when first was head and second is tail) = 1/2 * 1/2 = 1/4
Probability of getting Third Head ( such that first was tail and second was tail) = 1/2 * 1/2 * 12 = 1/8

As you can notice the probability will be 1/8 whatever the sequence of Heads and tails.
We can arrange 2 Tails and one head in 3 ways (3C2 ways OR (3P3)/2! OR enumeration, whichever method you prefer to get here)

Since the favorable even can happen in either of these 3 ways, so Probability = 3 * 1/8 = 3/8
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Re: John tossed a fair coin 3 times. What is the probability that the outc  [#permalink]

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New post 25 Sep 2016, 21:43
Total outcomes : 2^3 = 8

HHH
HHT
HTH
THH
TTH
THT
HTT
TTT

Exactly 2 tails are 3/8
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Re: John tossed a fair coin 3 times. What is the probability that the outc  [#permalink]

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New post 26 Sep 2016, 03:58
Bunuel wrote:
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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Re: John tossed a fair coin 3 times. What is the probability that the outc  [#permalink]

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New post 18 Oct 2018, 18:35
Bunuel wrote:
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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Re: John tossed a fair coin 3 times. What is the probability that the outc   [#permalink] 18 Oct 2018, 18:35
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