Set = {13, 14, ...., 30}

Set = {13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30}

Number of samples = 18

Prime numbers are numbers with only two factors (1 and itself). Thus only 1 and the number itself can divide the number without a remainder.

Set of prime numbers from the set = {13, 17, 19, 23, 29}

Number of prime numbers = 5

P(Prime) = $\frac{5}{18}$

P(Both fail) = P(John fail) and P(James fail)

And means multiplication (x)

P(Both fail) = P(John fail) x P(James fail)

Probability of not = 1 - Probability of is

P(John pass) = $\frac{3}{4}$

P(John fail) = 1 - P(John pass)

P(John fail) = 1 - $\frac{3}{4}$

Every number is being divided by 1.

1 = $\frac{1}{1}$

P(John fail) = $\frac{1}{1}$ - $\frac{3}{4}$

P(John fail) = $\frac{\mathrm{4\; x\; 1\; -\; 1\; x\; 3}}{4}$

P(John fail) = $\frac{\mathrm{4\; -\; 3}}{4}$

P(John fail) = $\frac{1}{4}$

P(James pass) = $\frac{3}{5}$

P(James fail) = 1 - P(James pass)

P(James fail) = 1 - $\frac{3}{5}$

P(James fail) = $\frac{1}{1}$ - $\frac{3}{5}$

P(James fail) = $\frac{\mathrm{5\; x\; 1\; -\; 1\; x\; 3}}{5}$

P(James fail) = $\frac{\mathrm{5\; -\; 3}}{5}$

P(James fail) = $\frac{2}{5}$

P(Both fail) = P(John fail) x P(James fail)

P(Both fail) = $\frac{1}{4}$ x $\frac{2}{5}$ = $\frac{1}{10}$