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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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    $\begingroup$ Welcome to MathSE. When you pose a question here, it is expected that you include your own thoughts on the problem. Please indicate what you have tried and where you are stuck so that you receive responses appropriate to your skill level. $\endgroup$ – N. F. Taussig Jan 16 '17 at 13:59
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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).]

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  • $\begingroup$ e)probablity of selecting 1st student=35/35 probablity of 2nd undergrad=11/34 combine=1*11/34=11/34 $\endgroup$ – The predictor Jan 17 '17 at 8:41
  • $\begingroup$ 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. $\endgroup$ – BruceET Jan 18 '17 at 5:40
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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) = $\frac{P(A \cap B)}{P(B)}$

Where P(B) is undergraduate student.

d. $P(U|G) = \frac{P(G \cap U)}{P(G)}$

Then,

$P(G \cap U) = P(U|G) \cdot P(G)$

e. $P(U) = P(G \cap U) + P(U_1 \cap U_2)$

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  • $\begingroup$ i m stuck in c d and e ..plz help $\endgroup$ – The predictor Jan 16 '17 at 15:26
  • $\begingroup$ Then add a and b solution to show what you have tried $\endgroup$ – Kanwaljit Singh Jan 16 '17 at 16:27
  • $\begingroup$ If any doubt ask me. $\endgroup$ – Kanwaljit Singh Jan 16 '17 at 16:51
  • $\begingroup$ I appreciate that you are trying to give clues without working the problem, but your clues for parts (d) and (e) are not helpful. (Rather than down-voting, I have edited your Answer to shade these parts. Please re-edit or remove as appropriate.) $\endgroup$ – BruceET Jan 17 '17 at 0:40

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