Question

The probability that a student pilot passes the
written test for a private pilot’s license is 0.7. Find the
probability that a given student will pass the test
(a) on the third try;
(b) before the fourth try.

Answer 1

Answer on Question #72701, Math / Statistics and Probability

The probability that a student pilot passes the written test for a private pilot's license is 0.7. Find the probability that a given student will pass the test

(a) on the third try;

(b) before the fourth try.

Solution

Let us consider that a written test for a pilot's license is a trial. Trials are independent. If a student passes the test, we consider that as a success.

Since the person needs to pass the test only once, we are looking for the number of trials to obtain one success, i.e. the Geometric distribution.

Let $X$ be the random variable representing the number of tries to pass the test, that is $X$ is the number of trials required to get $1^{st}$ success. This is a negative binomial or geometric variable.

$p = P(success) = 0.7$ and the number of successes to observe $r = 1$.

$$\begin{array}{l} P (X = x) = b ^ {*} (x; r, p) \text { for } x = r, r + 1, r + 2, \dots \\ P (X = x) = \left( \begin{array}{c} x - 1 \\ r - 1 \end{array} \right) p ^ {r} q ^ {x - r} \\ \end{array}$$

The geometric distribution is a special case of the negative binomial distribution such that $r = 1$.

(a) on the third try

The probability $P$ that the student passes on the third try

$$P (X = 3) = \binom {3 - 1} {1 - 1} (0.7)^{1} (1 - 0.7)^{3 - 1} = 0.063$$

(b) before the fourth try

$$\begin{array}{l} P (X \leq 3) = p + p (1 - p) + p (1 - p)^{2} \\ P (X \leq 3) = 0.7 + 0.7 (1 - 0.7) + 0.7 (1 - 0.7)^{2} = 0.973 \\ \end{array}$$

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