Hello! I'd be happy to help you with this probability question.
(a) To find the probability of obtaining at least one six when throwing a cubic die 4 times, we can calculate the probability of not getting a six in any single throw and then subtract that from 1 (which gives us the probability of getting at least one six).
The probability of not getting a six on a single throw is 5/6 (since there are 6 outcomes and only 1 is a six). So, the probability of not getting a six in 4 throws is (5/6)^4. Therefore, the probability of getting at least one six in 4 throws is 1 - (5/6)^4.
(b) To find the probability of getting all odd numbers when throwing the die 4 times, we need to calculate the probability of getting an odd number on a single throw, which is 3/6 (since there are 3 odd numbers out of 6 outcomes). Therefore, the probability of getting all odd numbers in 4 throws is (3/6)^4.
(c) Finally, to find the probability of getting at least one even number in 4 throws, we can use a similar approach as in part (a). The probability of not getting an even number on a single throw is 3/6 (since there are 3 odd numbers out of 6 outcomes). So, the probability of not getting an even number in 4 throws is (3/6)^4. Therefore, the probability of getting at least one even number in 4 throws is 1 - (3/6)^4.
I hope this helps! Let me know if you need further clarification.
