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

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by GMATGuruNY » Sat Oct 06, 2018 6:58 am
AAPL wrote:EMPOWERgmat

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
P(TTH) = 1/2 * 1/2 * 1/2 = 1/8.
Since H can occur on the 1st, 2nd or 3rd flip, we multiply by 3:
3*(1/8) = 3/8.

The correct answer is C.
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by [email protected] » Sat Oct 06, 2018 1:49 pm
Hi All,

We're told that John tossed a fair coin 3 times. We're asked for the probability that the outcome was "tails" EXACTLY twice. This question can be approached in a number of different ways; sometimes the fastest way to answer these types of Probability questions is to just 'map out' the possibilities.

Since each coin toss has 2 possible outcomes and we're flipping 3 times, there are (2)(2)(2) = 8 possible outcomes. They are:

HHH
HHT
HTH
THH

TTT
TTH
THT
HTT

3 of the 8 possible outcomes have exactly 2 'tails.'

Final Answer: C

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by Scott@TargetTestPrep » Thu Oct 18, 2018 5:31 pm
AAPL wrote:EMPOWERgmat

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

by fskilnik@GMATH » Fri Oct 19, 2018 6:09 am
AAPL wrote:EMPOWERgmat

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
$$? = P\left( {\,{\rm{exact}}\,{\rm{2}}\,\,{\rm{tails}}\,{\rm{in}}\,{\rm{3}}\,{\rm{tosses}}\,} \right) = {{C\left( {3,2} \right)} \over {2 \cdot 2 \cdot 2}} = {3 \over 8}$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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