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A balanced coin is tossed four times. List the possible outcomes and compute the probability of each of the three events: (a) exactly three heads (b) at least one head (c) the number of heads equals the number of tails (d) the number of heads exceeds the number of tails

My solution:

A balanced coin is tossed four times, So the possible outcomes can be following:

possible outcomes are 2 and trials are 4

Sample space is determined by= $2^4=16$

HHHH HTHH THHH HTHT

HHHT HTTH TTHH THTH

HHTT HHTH TTTH THHT

HTTT TTTT TTHT THTT

a) Probability of exactly three heads$=4/16=1/4$ b) Probability of At least one head$=15/16$ c) Probability of that the no. of heads equals the no. of tails$=6/16$ d) Probability of that the no. of heads exceeds the no. of tails$=4/16=1/4$

Could you see it for me is it correct?

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  • $\begingroup$ d is incorrect. P(H>T) = P(H=3)+P(H=4) = 5/16. $\endgroup$
    – Doug M
    Commented May 19, 2016 at 4:18
  • $\begingroup$ @shaihorowitz It was't there when I put in mine. Why so defensive? $\endgroup$
    – Doug M
    Commented May 19, 2016 at 4:30

2 Answers 2

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The first $3$ answers are correct, but the $4$th answer is wrong:

  • Probability of exactly $3$ heads $=\frac{\binom43}{2^4}=\frac{4}{16}$
  • Probability of at least $1$ head $=\sum\limits_{n=1}^{4}\frac{\binom4n}{2^4}=\frac{15}{16}$
  • Probability that the no. of heads equals the no. of tails $=\frac{\binom42}{2^4}=\frac{6}{16}$
  • Probability that the no. of heads exceeds the no. of tails $=\sum\limits_{n=3}^{4}\frac{\binom4n}{2^4}=\frac{5}{16}$
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Your correct for the most part. Slight error in part d. Consider the Probability of that the no. of heads exceeds the no. of tails = probability of 3 or more heads = probability of 3 heads plus probability of 4 heads

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