H2 Maths 6.2 - Binomial Distribution: P(X = r), E(X), Var(X)

Step 1 - Set up the binomial model

A firm interviews twelve candidates. Each candidate independently has a zero point three probability of accepting an offer.

Step 1 - Set up the binomial model

Let X be the number who accept.

Step 1 - Set up the binomial model

We check the four binomial conditions.

Step 1 - Set up the binomial model

Fixed number of trials: yes, n equals twelve.

Step 1 - Set up the binomial model

Constant probability of success: yes, p equals zero point three.

Step 1 - Set up the binomial model

Independent trials: yes, each candidate decides independently.

Step 1 - Set up the binomial model

Two outcomes per trial: accept or reject.

Step 1 - Set up the binomial model

So X follows a binomial distribution with n equals twelve and p equals zero point three.

Step 2 - Find P(X = 4)

P of X equals four uses the binomial formula.

Step 2 - Find P(X = 4)

Twelve choose four, times zero point three to the power four, times zero point seven to the power eight.

Step 2 - Find P(X = 4)

Twelve choose four is four hundred and ninety-five.

Step 2 - Find P(X = 4)

Zero point three to the four is zero point zero zero eight one.

Step 2 - Find P(X = 4)

Zero point seven to the eight is approximately zero point zero five seven six.

Step 2 - Find P(X = 4)

Multiplying these gives approximately zero point two three one.

Step 3 - Find P(X ≥ 2) using complement

For at least two, use the complement.

Step 3 - Find P(X ≥ 2) using complement

P of X greater than or equal to two equals one minus P of X equals zero minus P of X equals one.

Step 3 - Find P(X ≥ 2) using complement

P of X equals zero is zero point seven to the twelve, which is approximately zero point zero one three eight.

Step 3 - Find P(X ≥ 2) using complement

P of X equals one is twelve times zero point three times zero point seven to the eleven, approximately zero point zero seven one two.

Step 3 - Find P(X ≥ 2) using complement

So one minus zero point zero one three eight minus zero point zero seven one two gives approximately zero point nine one five.

Step 4 - Compute E(X) and Var(X)

For a binomial distribution, the shortcuts are simple.

Step 4 - Compute E(X) and Var(X)

E of X equals n p, which is twelve times zero point three, giving three point six.

Step 4 - Compute E(X) and Var(X)

Variance of X equals n p times one minus p.

Step 4 - Compute E(X) and Var(X)

That is twelve times zero point three times zero point seven, which equals two point five two.

Step 4 - Compute E(X) and Var(X)

On average, about three to four candidates accept, with a standard deviation of roughly one point five nine.