Explanation:
we have 2 probabilities :
1. that a driver is wearing their safety belts : 0.68
2. that the driver is NOT wearing the safety belts :
1-0.68 = 0.32
A.
3 drivers are wearing it, 2 are not is a combination of
0.68³×0.32²
and how many combinations of these 5 probabilities can we have (e.g. the first 3 are wearing them, or the last 3 are wearing them, or ...) ?
that is 5 over 3 combinations :
5! / (3! × (5-3)!) = 5! / (3! × 2!) = 5×4/2 = 5×2 = 10
so, we have
10×0.68³×0.32² = 0.321978368
B
I am not sure I understand the question.
I assume this means from the 5 drivers they pull over the first 4 don't wear the seat belts, and only the fifth does.
so that is then
0.32⁴×0.68
and only one combination is possible. so, the result is just the product of these probabilities :
0.32⁴×0.68 = 0.0071303168