Assume that the probability a child is a boy is 0.51 and...

Assume that the probability a child is a boy is 0.51 and that the sexes of children born into a family are
independent. What is the probability that a family of five children has
a) exactly three boys?
b) at least one boy?
c) at least one girl?
d) all children of the same sex?

Key Concepts

Combinatorics

The branch of mathematics concerned with counting and arranging objects. In probability, combinatorial methods, such as calculating binomial coefficients, are used to determine the number of ways a particular outcome can occur.

Complement Rule

A strategy in probability that involves calculating the probability of an event by subtracting the probability of its complement (the event not occurring) from one. This rule simplifies finding probabilities, especially in cases like 'at least one' scenarios.

Binomial Distribution

A probability distribution that describes the number of successes in a fixed number of independent trials, where each trial has the same probability of success. It is essential for calculating probabilities of obtaining a specific number of outcomes in repeated experiments.

Independent Events

A fundamental concept in probability theory where the outcome of one event does not affect the outcome of another. This assumption enables the use of the binomial distribution when modeling repeated trials.

Transcript

00:01 Okay, so given that, the probability that a child would be a boy is 0 .51, so just greater than 50%, then the probability of having a girl, or that's just going to be 1 minus 0 .51, which is 0 .49.

00:24 So here, let's let p be the probability of the child being a boy.

00:31 So we have that p is equal to 0 .51.

00:37 Therefore, q is going to be equal to 1 minus 0 .51, or q is equal to 0 .49.

00:48 Okay? and we're given that the family has five children.

00:53 So we're going to be using the theorem here that the probability of exactly k successes in n independent for newly tribes.

01:02 Such that the probability of successes being p and the probability of failure being q which is equal to one minus p we have that c of n k uh p to the k to the k q to the n minus k so um part a as any of the three children can be boys there will would be c -5 -3 possible ways.

01:44 Since the five children are independent of each other, so the probability of exactly three successes in five independent for newly trials, with probability of success, 0 .11, and probability of failure, 0 .49, we have that c -53 times c .0 .51 cubed.

02:09 Times 0 .49 squared, which gives us 0 .3184.

02:25 Okay.

02:27 And for b, well, we have defined the probability here that there is at least one boy.

02:34 So that is that there may be one or two or three or four or possibly all five boys.

02:41 So it's enough here if we deduct the probability that all five children are girls from the total probability, which is one.

02:50 So the chance that all the children are girls in five independent for newly trials with our success being 0 .49 and the failure now being 0 .51 would be c of 55, which is 0 .45, which is 0 .4.

03:12 49 to the fifth.

03:17 So we get here, this is basically saying 1 minus 0 .49 to the 5th, which is equal to 0 .9717...

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