Using Conditional Probability and Product Rules

A class contains 35 students: 11 undergrads and 24 grad students. Of the undergraduates, 4 are female and 7 and male. Of the grad students, 5 are female and 19 are male.

a)I randomly select a student from the class. What is the probability the student is female?

b) I randomly select a student from the class. Given that the student I select is an undergraduate, what is the conditional probability that they are male?

c)I randomly select a student from the class. Given that the student I select is a male,what is the conditional probability that they are an undergraduate?

d)I randomly select two students from the class, without replacement, in order. Given that the first student I select is a grad student, what is the conditional probability the second student I select is an undergraduate?

e)I randomly select two students from the class, without replacement, in order. What is the (unconditional) probability the second student I select is an undergraduate?

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Here is some help with parts (d) and (e):

You should use what you find in your text together with @KanwaljitSingh's Comments to get started with (a)-(c). I suggest you edit your Question, adding numerical answers after each of these parts, as you find them (along with a few words of explanation). Otherwise this Question may be "Closed" for lack of engagement before you get all the help you need.

(d) $P(G_1 \cap U_2) = P(G_1)P(U_2|G_1) = (24/35)(11/34) = ??$

(e) $P(U_2) = P(G_1 \cap U_2) + P(U_1 \cap U_2).$ You have already found the first probability. Use a similar method to find the second and add. [Hint: By symmetry, $P(U_2) = P(U_1) = 11/35.$ You can use this to check your answer to (e).]

• e)probablity of selecting 1st student=35/35 probablity of 2nd undergrad=11/34 combine=1*11/34=11/34 – The predictor Jan 17 '17 at 8:41
• No: For (3) it's (24/35)*(11/34)+(11/35)*(10/34) = 0.3142857 = 11/35. Sorry for my typo with 11/36 instead of 11/35, now fixed. – BruceET Jan 18 '17 at 5:40

Hint - Use following formulas.

1. Probability = $\frac{\text{Favourable Cases}}{\text{Total Cases}}$

2. Conditional Probability P(A|B) = $\frac{P(A \cap B)}{P(B)}$

3. Unconditional Probability P(A) = $\frac{\text{number of times independent event occurs}}{\text{total number of possible outcomes}}$

Edit -

c. P(A|B) = $\frac{P(A \cap B)}{P(B)}$

d. $P(U|G) = \frac{P(G \cap U)}{P(G)}$
$P(G \cap U) = P(U|G) \cdot P(G)$
e. $P(U) = P(G \cap U) + P(U_1 \cap U_2)$