Question

10) Thirty-seven percent of the American population has blood type O+. What is the probability that at least four of the next five Americans tested will have blood type O+?

Answer 1

Required probability = 0.066

Explanation:

We are given that Thirty-seven percent of the American population has blood type O+.

Firstly, the binomial probability is given by;

`P(X=r) =\binom{n}{r}p^(r)(1-p)^(n-r) for x = 0,1,2,3,....`

where, n = number of trails(samples) taken = 5 Americans

r = number of successes = at least four

p = probability of success and success in our question is % of

the American population having blood type O+ , i.e. 37%.

Let X = Number of people tested having blood type O+

So, X ~
`Binom(n=5,p=0.37)`

So, probability that at least four of the next five Americans tested will have blood type O+ = P(X >= 4)

P(X >= 4) = P(X = 4) + P(X = 5)

=
`\binom{5}{4}0.37^(4)(1-0.37)^(5-4) + \binom{5}{5}0.37^(5)(1-0.37)^(5-5)`

=
`5*0.37^(4)*0.63^(1) +1*0.37^(5)*1` = 0.066.

Answer 2

0.06597

Explanation:

Given that thirty-seven percent of the American population has blood type O+

Five Americans are tested for blood group.

Assuming these five Americans are not related, we can say that each person is independent of the other to have O+ blood group.

Also probability of any one having this blood group = p = 0.37

So X no of Americans out of five who were having this blood group is binomial with p =0.37 and n =5

Required probability

=The probability that at least four of the next five Americans tested will have blood type O+

=
`P(X\geq 4)\\= P(X=4)+P(x=5)\\= 5C4 (0.37)^4 (1-0.37) + 5C5 (0.37)^5\\= 0.06597`