Question

There are 4 blood types, and not all are equally likely to be in blood banks. 49% of donations are type O blood, 27% of donations are type A blood, 20% of donations are type B blood, and 4% of donations are type AB blood. A person with type A blood can safely receive blood transfusions of type O and type A blood. If three donations are made, what is the probability that at least 1 of them can be safely used in a blood transfusion on someone with type A blood? 0.0138 0.4390 0.5610 0.9862

Answer 1

Answer:

0.9862.

Step-by-step explanation:

For each donation, there are only two possible outcomes. Either it can be safely used in a blood transfusion on someone with type A blood, or it cannot. The probability of a donation being safely used is independent of other donations, which means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Type A blood:

Can receive from type O(49%) and from type A(27%), which means that:

$p = 0.49 + 0.27 = 0.76$

If three donations are made, what is the probability that at least 1 of them can be safely used in a blood transfusion on someone with type A blood?

Three donations mean that $n = 3$

This probability is:

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{3,0}.(0.76)^{0}.(0.24)^{3} = 0.0138$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0138 = 0.9862$

The answer is 0.9862.

Answer 2

Answer:

b

Step-by-step explanation: