The question reads as follows: "During a party attended by 3 females and 3 males, 3 people at random enter a previously empty room. What is the probability that there are exactly 2 males in the room?"
I can do the first half of the problem without any issues. I simply plug the numbers into the combination formula — c = (6!) / [(3)!(6-3)!]. This resolves as c = 20, so I know that there are twenty possible combinations.
This is where things get tricky — how do I find the number of the those combinations which have two males and one female? My SAT book suggests that I write out every combination, but I don't like this approach. I know there must be a "more mathematical" way to derive this answer (9).
The probability (i.e., correct answer) given is 9/20.
Thanks so much for your help!
Answers & Comments
Verified answer
Find the probability of the order being - male, male, female.
This ends up being 0.15 from the calculation (3/6) x (2/5) x (3/4).
Multiply that value by three for the three ways to get two males in the room - M/M/F, M/F/M, F/M/M.
You get 0.45, or 9/20.
Possible anwsers
M M F
M F M
F M M
when two probabilities independent according to law of independent? multiply
lets go with option 1 M M F
Chance of picking male first try 3/6=person selected/total ppl
2nd= 2/5 because one person already picked out
3rd must account for woman type since we haven't picked them 3/4 is their total chance of being picked
3(3/6 times 2/5 times 3/4)
Times 3 because this can happen in 3 different ways as mentioned in p1.
Hi Angie,
This question is WAY too hard for the SAT. I've studied hundreds of real SATs, and I've never seen a probability question this difficult.