# Conditional Probability and Independent Events

Assume that there are equal numbers of males and females at a school. The probability is 1/5 that a male student and 1/20 that a female student will be taking a science course. What is the probability that:

a.) a randomly selected student will be a male science student b.) a randomly selected student will be a science student c.) a student is a science student (S), given that she is female (F)

I know that Male taking science is 1/5. So, male not taking science is 4/5. Female taking science is 1/20 so female not taking science is 19/20.

However, I am confused how to set up the problem.

Thank you again for the help!

• You need to check the wording of your question since it would appear that the answer to part c) of your question is given in the data, i.e. the probability they are a science student given that they are female is $\frac {1}{20}$ – David Quinn Oct 1 '17 at 20:02
• Yes, the question is worded very strangely. I thought the answer to c) would have been 1/20. – Megan C. Oct 1 '17 at 20:07
• But I am still not sure how to solve the rest of this problem. Thank you! – Megan C. Oct 1 '17 at 20:37
• Have you tried drawing a tree diagram? – David Quinn Oct 1 '17 at 21:30

I believe the awkward wording is suggesting a direct partition: $$P(\text{Male},\text{Science})+P(\text{Female},\text{Science})+(\text{Male},\text{Not Science})+P(\text{Female},\text{Not Science}) = 1$$ Otherwise there would be the word "given" explicitly.

1. $P(\text{Male},\text{Science}) = 1/5$

2. $P(\text{Male},\text{Science})+P(\text{Female},\text{Science}) = P(\text{Science}) = 1/20 + 1/5 = 1/4$

3. $P(\text{Science} | \text{Female}) = \frac{P(\text{Female} | \text{Science})*P(\text{Science})}{P(\text{Female})} = \frac{P(\text{Female}, \text{Science})}{P(\text{Female})} = \frac{1/5}{1/2} = 1/10$

• You have not taken into account the probability that a given student is male or female. – N. F. Taussig Oct 2 '17 at 12:58

What is the probability that a randomly selected student is a male science student?

Let $P(M)$ denote the probability that a student is male. Let $P(S)$ denote the probability that a student is a science student. Let $P(M \cap S)$ denote the probability that a student is a male science student. Let $P(S \mid M)$ denote the probability that a student is a science student given that the student is male. The probability that a randomly selected student is a male science student is $$P(M \cap S) = P(S \mid M)P(M) = \frac{1}{5} \cdot \frac{1}{2} = \frac{1}{10}$$

What is the probability that a randomly selected student is a science student.

Let $P(M)$, $P(S)$, $P(M \cap S)$, and $P(S \mid M)$ be defined as above. Let $P(F)$ be the probability that a student is female. Let $P(F \cap S)$ be the probability that a student is a female science student. Let $P(S \mid F)$ be the probability that a student is a science student given that she is female. Assuming no students are intersex, the probability that a student is a science student is $$P(S) = P(F \cap S) + P(M \cap S)$$ which you can compute as follows: $$P(S) = P(F \cap S) + P(M \cap S) = P(S \mid F)P(F) + P(S \mid M)P(M)$$

What is the probability that a student is a science student given that she is female?

The probability that a student is a science student given that she is female is $P(S \mid F)$, which was given in the statement of the problem.

The probability that a student is female given that the student is a science student is a more interesting question. That is $$P(F \mid S) = \frac{P(F \cap S)}{P(S)} = \frac{P(S \mid F)P(F)}{P(S \mid F)P(F) + P(S \mid M)P(M)}$$

I will go very basic

Say there are 100 Males and 100 Female students So 1/5 MS = 20/100 MS and 1/20 FS= 5/100FS

a -> Random MS prob => total 200 Students and 20 MS. So 20/200 = 1/10.

 In formula it is : P(MS) = P(M).P(S/M) = (1/2).(1/5) = 1/10


b -> Random S prob => total 200 and 20MS+5FS. so 25/200 = 1/8

 In formula : P(S) = P(FS)+P(MS) =  (1/2).(1/5)+(1/2).(1/20) = 1/8