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

Answer:

0.06597

Step-by-step 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+

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

Answer 2

Answer:

Required probability = 0.066

Step-by-step 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.