FS1-3 Geometric and negative binomial distributions — printable coverage pack

FS1-3.1 · Geometric and negative binomial distributions.

For independent Bernoulli trials with success probability $p$ , a geometric variable $X$ counts the trial of the first success: $P(X=x)=p(1-p)^{x-1}$ for $x=1,2,\ldots$ .
A negative binomial variable $Y$ counting the trial of the $r$ th success has $P(Y=y)=\binom{y-1}{r-1}p^r(1-p)^{y-r}$ for $y=r,r+1,\ldots$ .
The combination $\binom{y-1}{r-1}$ places the first $r-1$ successes before the final, fixed success on trial $y$ .
A common error is to use $\binom{y}{r}$ , forgetting that the last trial must be a success.

1. Independent trials have success probability $0.3$ . Find the probability that the first success occurs on trial $4$ .[2 marks]
Answer
- $0.1029$
Method: The first three trials must fail and trial $4$ must succeed, so the probability is $(0.7)^3(0.3)=0.1029$ .

1. Independent trials have success probability $0.4$ . Find the probability that the third success occurs on trial $5$ .[3 marks]
Answer
- $0.13824$
Method: Exactly two of the first four trials must be successes, followed by a success. Hence $P(Y=5)=\binom{4}{2}(0.4)^3(0.6)^2=0.13824$ .

1. Each route attempt succeeds independently with probability $0.35$ . Calculate the probability that the second successful route is completed by the fifth attempt.[4 marks]
Answer
- $0.571585$
Method: The second success occurs by attempt $5$ exactly when there are at least two successes among the first five attempts. Its complement has zero or one success, so the probability is $1-(0.65)^5-\binom51(0.35)(0.65)^4=0.571585$ .

For a geometric variable that counts the trial of the first success, $E(X)=1/p$ and $\operatorname{Var}(X)=(1-p)/p^2$ .
The standard deviation is $\sqrt{1-p}/p$ , obtained by taking the positive square root of the variance.
For example, when $p=0.25$ , the mean is $4$ trials and the variance is $12$ .
A common error is to use the alternative convention that counts failures before the first success; the Edexcel FS₁ convention starts at $X=1$ .

1. The random variable $X$ is geometric with parameter $p=0.2$ and counts the trial of the first success. Find its mean and variance.[2 marks]
Answer
- $E(X)=5$
- $\operatorname{Var}(X)=20$
Method: $E(X)=1/p=1/0.2=5$ and $\operatorname{Var}(X)=(1-p)/p^2=0.8/0.2^2=20$ .

If $X$ counts the trial of the $r$ th success, then $E(X)=r/p$ and $\operatorname{Var}(X)=r(1-p)/p^2$ .
The ratio $\operatorname{Var}(X)/E(X)=(1-p)/p$ is useful when the mean and variance are given and $p$ is unknown.
For example, $r=3$ and $p=0.6$ give mean $5$ and variance $10/3$ .
A common error is to use the formulas for the number of failures, whose mean is $r(1-p)/p$ ; FS₁ defines $X$ as the trial number of the $r$ th success.

1. The random variable $X$ counts the trial on which the fourth success occurs, with success probability $0.25$ . Find $E(X)$ and $\operatorname{Var}(X)$ .[2 marks]
Answer
- $E(X)=16$
- $\operatorname{Var}(X)=48$
Method: Here $r=4$ and $p=0.25$ . Thus $E(X)=r/p=4/0.25=16$ and $\operatorname{Var}(X)=r(1-p)/p^2=4(0.75)/0.25^2=48$ .

1. Successive trials are independent with success probability $0.4$ . Let $X$ be the trial on which the sixth success occurs. Find the probability that $X\leq8$ .[5 marks]
Answer
- $P(X\leq8)=0.04981$ to $4$ significant figures
Method: Sum the negative binomial probabilities for $x=6,7,8$ : $P(X\leq8)=\sum_{x=6}^{8}\binom{x-1}{5}(0.4)^6(0.6)^{x-6}=0.04980736\ldots$ .