Respuesta :
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+
= [tex]P(X\geq 4)\\= P(X=4)+P(x=5)\\= 5C4 (0.37)^4 (1-0.37) + 5C5 (0.37)^5\\= 0.06597[/tex]
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;
[tex]P(X=r) =\binom{n}{r}p^{r}(1-p)^{n-r} for x = 0,1,2,3,....[/tex]
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 ~ [tex]Binom(n=5,p=0.37)[/tex]
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)
= [tex]\binom{5}{4}0.37^{4}(1-0.37)^{5-4} + \binom{5}{5}0.37^{5}(1-0.37)^{5-5}[/tex]
= [tex]5*0.37^{4}*0.63^{1} +1*0.37^{5}*1[/tex] = 0.066.
