# A balanced coin is tossed four times.

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?

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

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 $=\sum\limits_{n=3}^{4}\frac{\binom4n}{2^4}=\frac{5}{16}$