We want to find the probability that the two students chosen for the duet are boys. We will find that the probability that both students chosen for the duet are boys is 0.458
If we assume that the selection is totally random, then all the students have the same<em> </em><em>probability </em><em>of being chosen.</em>
This means that, for the first place in the duet, the probability of randomly selecting a boy is equal to the quotient between the number of boys and the total number of students, this is:
P = 11/16
For the second member of the duet we compute the probability in the same way, but this time there is one student less and one boy less (because one was already selected).
Q = 10/15
The joint probability (so both of these events happen together) is just the product of the individual probabilities, this will give:
Probability = P*Q = (11/16)*(10/15) = 0.458
So the probability that both students chosen for the duet are boys is 0.458
If you want to learn more, you can read:
brainly.com/question/1349408
Answer:
Step-by-step explanation:
Given that the number of children in a household has a binomial distribution with parameters n-8 and p-50%
As per binomial definition we have
=
a) P(x=3 or 7) = P(3)+P(7) = 0.2188+0.0313=0.2501
b) P(X≤5)=0.8555
c) P(X≥3)=0.8555
d) P(x<7) =0.9648
