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Alex tosses a coin four times. On two of the tosses (we

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by BTGmoderatorLU » Sun Aug 04, 2019 7:20 am

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Source: Veritas Prep

Alex tosses a coin four times. On two of the tosses (we don't know which two), he gets 'Heads'. What is the probability that he gets 'Tails' on other two tosses?

A. \(\frac{1}{4}\)
B. \(\frac{3}{8}\)
C. \(\frac{1}{2}\)
D. \(\frac{6}{11}\)
E. \(\frac{3}{5}\)

The OA is D
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by Jay@ManhattanReview » Sun Aug 04, 2019 9:23 pm
BTGmoderatorLU wrote:Source: Veritas Prep

Alex tosses a coin four times. On two of the tosses (we don't know which two), he gets 'Heads'. What is the probability that he gets 'Tails' on other two tosses?

A. \(\frac{1}{4}\)
B. \(\frac{3}{8}\)
C. \(\frac{1}{2}\)
D. \(\frac{6}{11}\)
E. \(\frac{3}{5}\)

The OA is D
Note that this is a question on conditional probability. We know that the first two tosses have already occurred and we know their outcomes. Thus, after the 3rd and the 4th toss, we would have: {HHTT}, {HHHH}, {HHHT}, {HHTH} => at least 2 heads. So, our sample size is p(H ≥ 2)

p(H ≥ 2) = p(H = 2 & T = 2) + p(H = 3 & T = 1) + p(H = 4) = 4C2*[(1/2)^2]* [(1/2)^2] + 4C3*(1/2)^3* (1/2) + 4C4* (1/2)^4 = 11/16 : this is our denominator

Now since we want 2 Heads and 2 Tails, our numerator would be p(H = 2 & T = 2) = 4C2*[(1/2)^2]* [(1/2)^2] = (4.3)(1.2) / 2^4 = 3/8

Thus, the required probability = (3/8) / (11/16) = 6/11

The correct answer: D

Hope this helps!

-Jay
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by Scott@TargetTestPrep » Sun Aug 11, 2019 6:12 pm
BTGmoderatorLU wrote:Source: Veritas Prep

Alex tosses a coin four times. On two of the tosses (we don't know which two), he gets 'Heads'. What is the probability that he gets 'Tails' on other two tosses?

A. \(\frac{1}{4}\)
B. \(\frac{3}{8}\)
C. \(\frac{1}{2}\)
D. \(\frac{6}{11}\)
E. \(\frac{3}{5}\)

The OA is D
This conditional probability problem can be rephrased as follows:

If a coin is tossed four times and (at least) two of the tosses are heads, what is the probability that two tosses are heads and the other two tosses are tails?

Now, let's list the number of ways we have at least two heads in four tosses:

HHHH, HHHT, HHTH, HTHH, THHH, HHTT, TTHH, HTHT, THTH, HTTH, THHT

Of these 11 ways, we see that 6 consist of two heads and two tails (in bold); therefore, the probability is 6/11.

Answer: D

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[email protected]

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