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04 Aug 2017, 02:26
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
5%
(low)
Question Stats:
88% (01:20) correct
12%
(01:27)
wrong
based on 163
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If you are flipping the coin 4 times, what is the probability of landing on head two times?
A. \(\frac{3}{5}\)
B. \(\frac{4}{5}\)
C. \(\frac{3}{7}\)
D. \(\frac{4}{7}\)
E. \(\frac{3}{8}\)
A. \(\frac{3}{5}\)
B. \(\frac{4}{5}\)
C. \(\frac{3}{7}\)
D. \(\frac{4}{7}\)
E. \(\frac{3}{8}\)
sashiim20
Joined: 04 Dec 2015
Last visit: 05 Jun 2024
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Location: India
Concentration: Technology, Strategy
Schools: ISB '19 IIMB IIMA XLRI HEC Sept19 intake Rotman '21 NUS '21 NTU '20 Trinity MBA '20 Bocconi '21 Smurfit "21
WE:Information Technology (Consulting)
Schools: ISB '19 IIMB IIMA XLRI HEC Sept19 intake Rotman '21 NUS '21 NTU '20 Trinity MBA '20 Bocconi '21 Smurfit "21
Posts: 608
04 Aug 2017, 03:07
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MathRevolution
Coin is tossed \(4\) times.
P (\(2\) heads) \(= (\frac{1}{2} * \frac{1}{2} * \frac{1}{2} * \frac{1}{2} )* 4C2\)
P (\(2\) heads) \(= (\frac{1}{16}) * (\frac{4!}{2!2!})\)
P (\(2\) heads) \(= (\frac{1}{16}) *(\frac{4 * 3 * 2!}{2 * 1 * 2!})\)
P (\(2\) heads) \(= \frac{6}{16} = \frac{3}{8}\)
Answer (E)...
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General Discussion
04 Aug 2017, 04:41
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MathRevolution
Two results for each of four coin flips. When ways to perform tasks in series, we multiply. So that is 2×2×2×2 results in total. That is 2[square]4 or 16.
For the favourable case we need to count the ways to get 2 heads and 2 tails. The count of permutations of two pairs of symbols is: 4!/2!2!=6. This is easily confirmed by just counting.
|HHTT,HTHT,HTTH,THHT,THTH,TTHH}=6
Thus the probability is: 6/16, or:
0.375
