Answer:
Probability that two or more of them have Type A blood is 0.6242.
Step-by-step explanation:
We are given the approximate probabilities that a person will have blood type O, A, B, or AB.
<u>Blood Type</u> O A B AB
<u>Probability</u> 0.4 0.2 0.32 0.08
A group of 10 people are chosen randomly.
The above situation can be represented through Binomial distribution;
![P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....](https://tex.z-dn.net/?f=P%28X%3Dr%29%20%3D%20%5Cbinom%7Bn%7D%7Br%7Dp%5E%7Br%7D%20%281-p%29%5E%7Bn-r%7D%20%3B%20x%20%3D%200%2C1%2C2%2C3%2C.....)
where, n = number of trials (samples) taken = 10 people
r = number of success = two or more have Type A blood
p = probability of success which in our question is probability
that a person has Type A Blood, i.e; p = 20% or 0.20
<em>LET X = Number of person having Type A Blood</em>
So, it means X ~ Binom(n = 10, p = 0.20)
Now, Probability that two or more of them have Type A blood is given by = P(X
2)
P(X
2) = 1 - P(X = 0) - P(X = 1)
= ![1- \binom{10}{0}\times 0.20^{0} \times (1-0.20)^{10-0}-\binom{10}{1}\times 0.20^{1} \times (1-0.20)^{10-1}](https://tex.z-dn.net/?f=1-%20%5Cbinom%7B10%7D%7B0%7D%5Ctimes%200.20%5E%7B0%7D%20%5Ctimes%20%281-0.20%29%5E%7B10-0%7D-%5Cbinom%7B10%7D%7B1%7D%5Ctimes%200.20%5E%7B1%7D%20%5Ctimes%20%281-0.20%29%5E%7B10-1%7D)
=
= 0.6242
<em>Hence, the probability that two or more of them have Type A blood is 0.6242.</em>