Chapter 9, Problem 13RE

Solution

To state: The number of different outcomes that include at least two heads if a coin is tossed 5 times.

The resultant outcomes are 26.

Given information:

A coin is tossed 5 times to form different sequence of head and tail.

Explanation:

If a coin is tossed once, the number of outcomes is two (heads or tails).

Then the total number of outcomes if five coins are tossed: $2^5=32$

If a coin is tossed five times, the different outcomes of head will include one head, two head, three head, four head or five head.

The total number of outcomes that include one or no head: $C_{5_1}+C_{5_0}$

  $\begin{array}{c}{C}_{5_1}+C_{5_0}=\frac{5!}{1!\left(5-1\right)!}+\frac{5!}{0!\left(5-0\right)!}\\ =\frac{5\cdot 4!}{1!4!}+\frac{5!}{0!5!}\\ =5+1\\ =6\end{array}$

Now the total number of outcomes that include at least two heads will be equal to the total number of outcomes minus the outcomes that include one or no heads.

  $32-6=26$

Therefore, there are 26 outcomes that include at least two heads.