Question

A coin is flipped eight times where each flip comes up either heads or tails. How many possible outcomes a) are there in total? b) contain exactly three heads? c) contain at least three heads? d) contain the same number of heads and tails?
Rosen, Kenneth. Discrete Mathematics and Its Applications (p. 435). McGraw-Hill Higher Education. Kindle Edition.

Answer 1

Answer:

There are 256 ways in total.

There are 56 possible outcomes contain exactly three heads.

The possible outcomes contain at least three heads is 219.

The possible outcomes contain the same number of heads and tails are 70.

Step-by-step explanation:

Consider the provided information.

A coin is flipped eight times where each flip comes up either heads or tails.

Part (a) How many possible outcomes are there in total?

Each time we flip a coin it comes up either heads or tail.

Therefore the total number of ways are:

$2\times 2\times 2\times 2\times 2\times 2\times 2\times 2=2^8=256$

Hence, there are 256 ways in total.

Part (b) contain exactly three heads?

We want exactly 3 heads, therefore,

n=8 and r=3

According to the definition of combination: $\binom{n}{r}=\frac{n!}{r!(n-r)!}$

$\binom{8}{3}=\frac{8!}{3!(5)!}=56$

Hence, there are 56 possible outcomes contain exactly three heads.

Part (c) contain at least three heads?

For 3 heads: $\binom{8}{3}=\frac{8!}{3!(5)!}=56$

For 4 heads: $\binom{8}{4}=\frac{8!}{4!(4)!}=70$

For 5 heads: $\binom{8}{5}=\frac{8!}{5!(3)!}=56$

For 6 heads: $\binom{8}{6}=\frac{8!}{6!(2)!}=28$

For 7 heads: $\binom{8}{7}=\frac{8!}{7!(1)!}=8$

For 8 heads: $\binom{8}{8}=1$

Now add them as shown:

56+70+56+28+8+1=219

Hence, the possible outcomes contain at least three heads is 219.

Part (d) contain the same number of heads and tails?

Same number of heads and tails means that the value of r=4.

Therefore,

$\binom{8}{4}=\frac{8!}{4!(4)!}=70$

Hence, the possible outcomes contain the same number of heads and tails are 70.