i'm horrible in mathemantics...can some help with this question. find the probability that in FOUR tosses of a fair die a 3 appears
a)exactly no (zero)time
b)exactly three times
The probability of getting a 3 in a single roll is 1/6. The probabilty of not getting a 3 in a single roll is 5/6.
a) The probability of 'no 3' is (5/6)^4. You can get 'no 3' in only 1 way. So the probabilty is (1) (5/6)^4 = ......
b) The probability of 'exactly three 3's' is (1/6)^3 (5/6). You can get 'three 3's' in four different ways:
3, 3, 3, x
3, 3, x, 3
3, x, 3, 3
x, 3, 3, 3.
So the probability is (4) (1/6)^3 (5/6) = ......