Probability of student clearing given number of subjects

Question

A student appears in the examinations of four subjects B, C, P and M. Suppose that the probabilities of the student clearing examinations in these subjects are $\frac12; \frac13; \frac14$ and $\frac15$, respectively. Assuming that the performances of the student in four subjects are independent, find the following probabilities

1. clear all subjects

2. clear no subject

3. clear exactly one subject

4. clear exactly two subjects

5. clear atleast one subject

My try:

Let B, C, P and M denote the events that student clears the respective subjects

1. The required probability is $P(BCPM) = P(B)\cdot P(C)\cdot P(P)\cdot P(M) = \frac1{120}$ using independence

2. The required probability is $P(B^CC^CP^CM^C) = P(B^C)\cdot P(C^C)\cdot P(P^C)\cdot P(M^C)$ which can be calculated from given data

3. The required probability is $P(B^CCPM)+P(BC^CPM)+P(BCP^CM)+P(BCPM^C)$ which can again be computed easily using independence

4. The required probability is $P(B^CC^CPM)+P(B^CCP^CM)+P(BC^CP^CM)+P(B^CCPM^C)+P(BC^CPM^C)+P(BCP^CM^C)$ This will take time to compute, is there a different way to do this part?

5. The required probability is $1- P(\text{clear no subject})$ which can be computed using 2.

Is this correct?