A class group carried out a study of the makes and fuel types of cars in a large car park - Leaving Cert Mathematics - Question 6 - 2020

Given:

• Probability that a person passes: $p = \frac{1}{5}$
• Number of test takers: $n = 6$

To find the probability that Joe passes along with exactly 2 others, we can use the binomial probability formula:

$P(X = k) = \binom{n}{k}p^k(1-p)^{n-k}$

In this case,

• $k = 3$ (Joe + 2 others)
• $n = 6$ (total number of test takers)
• $p = \frac{1}{5}$
• $1 - p = \frac{4}{5}$

Thus:

$P(X = 3) = \binom{6}{3}\left(\frac{1}{5}\right)^3\left(\frac{4}{5}\right)^{3}$

Calculating:

$\binom{6}{3} = 20$

So,

$P(X = 3) = 20 \times \left(\frac{1}{5}\right)^3 \times \left(\frac{4}{5}\right)^{3} = 20 \times \frac{1}{125} \times \frac{64}{125} = \frac{1280}{3125}$

We need to express the probability of two or fewer people passing the test:

$P(\text{2 or less}) = P(0) + P(1) + P(2)$

Using the binomial formula:

1. For $k = 0$: $P(0) = \binom{n}{0} \left(\frac{1}{2}\right)^0 \left(\frac{1}{2}\right)^{n} = \left(\frac{1}{2}\right)^{n}$

2. For $k = 1$: $P(1) = \binom{n}{1} \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^{n-1} = n \left(\frac{1}{2}\right)^{n}$

3. For $k = 2$: $P(2) = \binom{n}{2} \left(\frac{1}{2}\right)^{2} \left(\frac{1}{2}\right)^{n-2} = \frac{n(n-1)}{2} \left(\frac{1}{2}\right)^{n}$

Summing these gives:

$P(\text{2 or less}) = \left(\frac{1}{2}\right)^{n} + n \left(\frac{1}{2}\right)^{n} + \frac{n(n-1)}{2} \left(\frac{1}{2}\right)^{n}$

Factoring out: $P(\text{2 or less}) = \left(\frac{1}{2}\right)^{n} \left(1 + n + \frac{n(n-1)}{2}\right)$

Consequently, the expression simplifies to: $\frac{am^2 + bn + c}{2^{n+1}}$

Identifying coefficients, we have:

• $a = 1$
• $b = 1$
• $c = 0$