Difficult Probability Q (1 Viewer)

[fuckit1991]

Question from JRAHS Trial 97...

Letters of the word JAMES are placed into a cup and letters of word RUSE placed into another cup. A letter is chosen at random from each cup.

What is the probability that two vowels are chosen given that at least one vowel is chosen?

 

[-tal-]

probability of 2 vowels chosen: 2/5 x 2/4 = 1/5

2 vowels in JAMES (2/5) , 2 in RUSE (2/4)

 

[ahhliss]

Uhm is that it? I don't get the part that goes "given that at least one vowel is chosen"

 

[fuckit1991]

^No, the answer is 2/7. I don't get the phrasing either, and this is the third time I've come across this type of question in JRAHS trial papers and I always get a wrong answer.

 

[Aerath]

Because you've given me the answer, I'm pretty sure I got it right. Unfortunately, my method is rather long.

 

[fuckit1991]

^Thanks, man.

 

[Aerath]

No problem. I'm sorry that I don't know a quicker method.

 

[3unitz]

No problem. I'm sorry that I don't know a quicker method.

total number of ways you can choose 2 letters = 20

P(both chosen) = (2/5)(2/4) = 4/20
P(none chosen) = (3/5)(2/4) = 6/20
P(at least 1 chosen) = 1 - (6/20) = 14/20

given at least 1 is chosen, there are now 14 possible outcomes, of which 4 lead to both being chosen. P(both being chosen) = 4/14 = 2/7

wouldnt say its much quicker for this question, but generally its probably a more faster method.

 

[Doctor Jolly]

Wow, you're good Aerath.

 

[ahhliss]

total number of ways you can choose 2 letters = 20

P(both chosen) = (2/5)(2/4) = 4/20
P(none chosen) = (3/5)(2/4) = 6/20
P(at least 1 chosen) = 1 - (6/20) = 14/20

given at least 1 is chosen, there are now 14 possible outcomes, of which 4 lead to both being chosen. P(both being chosen) = 4/14 = 2/7

wouldnt say its much quicker for this question, but generally its probably a more faster method.

I find this easier to understand ^^