An experiment consists of throwing two fair, standard, six-sided dice and noting the sum of the two numbers thrown - Leaving Cert Mathematics - Question 1 - 2015

To find the probability of a win:

• There are 10 combinations that result in a win (W) out of a total of 36 combinations when rolling two dice.

Thus, the probability of winning is given by:

$P(W) = \frac{10}{36} = \frac{5}{18}$

The probability of rolling a loss (L) is calculated by finding the number of loss outcomes.

• There are 26 combinations that result in a loss.

Thus, the probability of losing is:

$P(L) = \frac{26}{36} = \frac{13}{18}$

The probability that each of 3 successive throws results in a loss is:

$P(L,L,L) = \left(\frac{13}{18}\right)^3 \approx 0.3767$

To find this probability, we need to consider:

• The probability of winning (W) is (\frac{5}{18}) and the probability of losing (L) is (\frac{13}{18}).

The probability of exactly 2 wins in the first 9 throws can be calculated using the binomial formula:

$P(2 \text{ wins in } 9) = \binom{9}{2} \left(\frac{5}{18}\right)^2 \left(\frac{13}{18}\right)^7$

The third win occurring on the 10th throw contributes:

$P(3 \text{ wins, 3rd on 10th throw}) = P(2 \text{ wins in } 9) \times P(W) = \binom{9}{2} \left(\frac{5}{18}\right)^3 \left(\frac{13}{18}\right)^7$

Calculating this gives:

$P \approx 0.0791$