[Math] A balanced coin is tossed four times.

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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?

Best Answer

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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