If boys and girls are equally likely to be born, what is the probability that in a randomly selected family of 6 children, there will be at least one boy? (Find the answer using a formula. Round your answer to three decimal places.)
Binomial distribution
find probability of zero boys P(0)
answer is 1 - P(0)
n = 6
p = .5 and 1-p = .5
P(0) = C(6,0) (.5)^0 (.5^6)
coef for any n,0 = 1
so
P(0) = .5^6 = .0156
1-P(0) = .984
To find the probability that in a randomly selected family of 6 children there will be at least one boy, we can use the concept of complementary probability.
The complementary probability is the probability of the opposite event happening. In this case, the opposite event is the event of having no boys in the family.
If boys and girls are equally likely to be born, then the probability of having a boy in a single birth is 1/2, and the probability of having a girl is also 1/2.
To find the probability of having no boys in a family of 6 children, we need to find the probability of having only girls in all 6 births.
Since the outcomes of each birth are independent, we can multiply the probabilities together. Therefore, the probability of having only girls in all 6 births is (1/2)^6.
To find the probability of having at least one boy, we subtract the probability of having no boys from 1.
P(at least one boy) = 1 - P(no boys)
P(no boys) = (1/2)^6
P(at least one boy) = 1 - (1/2)^6 = 1 - 1/64 = 63/64
Therefore, the probability that in a randomly selected family of 6 children, there will be at least one boy is 63/64, which is approximately 0.984 (rounded to three decimal places).