In a population, the probability that a person has blue eyes is 0.7 - Leaving Cert Mathematics - Question 3 - 2019

The probability that one person has blue eyes is 0.7. Since the selections are independent, the probability that both people have blue eyes is:

$P( ext{both blue eyes}) = P( ext{blue eyes}) imes P( ext{blue eyes}) = 0.7 imes 0.7 = 0.49$

Thus, the probability that both have blue eyes is 0.49.

To find the probability that exactly two out of three chosen people have blue eyes, we can use the binomial probability formula:

P(X = k) = inom{n}{k} p^k (1 - p)^{n-k}

Where:

• $n = 3$ (total number of people)
• $k = 2$ (number of people with blue eyes)
• $p = 0.7$ (probability of having blue eyes)

Now substituting in:

1. The number of ways to choose 2 from 3 is given by: inom{3}{2} = 3

2. The probability calculation is: $P(X = 2) = 3 imes (0.7)^2 imes (0.3)^1$
   $= 3 imes 0.49 imes 0.3$ $= 3 imes 0.147 = 0.441$

Thus, the probability that exactly two of them have blue eyes is 0.441.

For the fourth person to be the only one with blue eyes, the first three must not have blue eyes, and the fourth must. We can calculate this as follows:

• The probability that the first three people do not have blue eyes is: $P( ext{not blue eyes})^3 = (0.3)^3 = 0.027$

• The probability that the fourth person has blue eyes is 0.7.

Thus, the total probability is:

$P = P( ext{first 3 not blue}) imes P( ext{fourth is blue}) = 0.027 imes 0.7 = 0.0189$

Therefore, the probability that the fourth person is the only one to have blue eyes is 0.0189.