If boys and girls are equally likely to be born, what is the probability that in a randomly selected family of 8 children, there will be at least one boy? (Find the answer using a formula. Give your answer correct to three decimal places.)
P( at least one boy) = 1-p(all girls)
1-(1/2)^ 8
1-1/256
(256-1)/256
= 255/256
= 0.996
To find the probability that in a randomly selected family of 8 children, there will be at least one boy, we can use the concept of the complement rule.
The complement rule states that the probability of an event occurring is equal to one minus the probability of the event not occurring. In this case, the event we are interested in is having at least one boy in a family of 8 children.
The probability that a family has no boys can be calculated by using the assumption that boys and girls are equally likely to be born. Since there are only two options for each child (boy or girl), the probability of having only girls in a family of 8 children is (1/2)^8, as both choices need to be girls for all 8 children.
Therefore, the probability of having at least one boy in a family of 8 children is 1 - (1/2)^8.
Now let's calculate this probability:
1 - (1/2)^8
= 1 - 1/256
= 255/256
So, the probability that in a randomly selected family of 8 children, there will be at least one boy is 0.996 (rounded to three decimal places).
To find the probability that in a randomly selected family of 8 children there will be at least one boy, we can use the concept of complementary probability.
The complementary probability is the probability that the event of interest does not occur. In this case, the event of interest is having at least one boy.
The probability of having all girls in a family of 8 children can be calculated as (1/2) * (1/2) * ... (1/2) = (1/2)^8 = 1/256.
Therefore, the complementary probability of having at least one boy is 1 - 1/256 = 255/256.
So, the probability that in a randomly selected family of 8 children, there will be at least one boy is 255/256.
Answer: 0.996