Kinda confused regarding these 2 qs. Thanks.
So for Q24. The answer is really similar to another question you have posted involving balls in a bag without replacement. You can consider the cases. What is involved here is multihypergeometric distribution.
If you understand the binomial distribution then you should be able to understand how to calculate the probability of a hypergeometric or multihypergeometric distribution.
The key idea is for both distribution
prob = num_events * prob_of_given_event