Respuesta :
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.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
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:
[tex]p = 0.49 + 0.27 = 0.76[/tex]
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 [tex]n = 3[/tex]
This probability is:
[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{3,0}.(0.76)^{0}.(0.24)^{3} = 0.0138[/tex]
[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0138 = 0.9862[/tex]
The answer is 0.9862.
The probability that at least 1 of them can be safely used in a blood is 0.9862.
Since, A person with type A blood can safely receive blood transfusions of type O and type A blood.
Given that,
49% of donations are type O blood
27% of donations are type A blood
So, [tex]p=0.49+0.27=0.76[/tex] and [tex]q=1-0.76=0.24[/tex]
Where p represent probability of success and q represent probability of unsuccessful.
Now use Binomial distribution,
[tex]P(X=r)=n_{C}_rp^{r} q^{n-r}[/tex]
Here three donations, it means that, n = 3
We have to find that the probability that at least 1 of them can be safely used in a blood.
So, [tex]P(X\geq 1)=1-P(X=0)[/tex]
[tex]P(X=0)=3_{C}_0(0.76)^{0} (0.24)^{3}=0.0138[/tex]
[tex]P(X\geq 1)=1-0.0138=0.9862[/tex]
Thus, The probability that at least 1 of them can be safely used in a blood is 0.9862.
Learn more about Binomial distribution here:
https://brainly.com/question/11388449
