(a)(i) Ans: \(\frac{1}{3}\)
There are 2 favorable outcomes (5 or 6) out of 6 possible cards. Thus, the probability is \(\frac{2}{6} = \frac{1}{3}\).
(a)(ii) Ans: 100
Multiply the probability \(\frac{1}{3}\) by the number of trials (300): \(\frac{1}{3} \times 300 = 100\).
(b)(i) Ans: \(\frac{2}{15}\)
Possible pairs summing to 5: (1,4), (2,3), (4,1), (3,2). There are 4 favorable outcomes out of \(6 \times 5 = 30\) total outcomes, giving \(\frac{4}{30} = \frac{2}{15}\).
(b)(ii) Ans: \(\frac{3}{5}\)
Square numbers in 1-6: 1, 4. Probability of neither card being square is \(\frac{4}{6} \times \frac{3}{5} = \frac{2}{5}\). Thus, the probability of at least one square is \(1 – \frac{2}{5} = \frac{3}{5}\).
