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Source: — Problem Solving |
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by Matt@VeritasPrep » Fri Sep 04, 2015 4:50 pm
All seven letters are different, so there are (7 choose 4) = 7! / 4!3! = 7*6*5 / 3*2 = 35 ways of picking four of the letters.

Only ONE of those 35 groups will give us what we want (the letters T, R, A, and M), so the probability is 1/35.
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by [email protected] » Fri Sep 04, 2015 4:53 pm
Hi tmontgomery,

What is the source of this question? I asked because the question is worded in an odd way, but I think the 'intent' of the question is the same as asking:

"There are 7 balls in a box - 3 red and 4 green. Balls will be randomly drawn one at a time and not replaced. What is the probability of selecting the 4 green balls on the first four draws?"

Under these conditions, we can deal with each individual 'event', then multiply the results.

Probability of drawing a green on the
first try: 4/7
second try: 3/6
third try: 2/5
fourth try: 1/4

(4/7)(3/6)(2/5)(1/4) = 1/35

Final Answer: C

GMAT assassins aren't born, they're made,
Rich
Last edited by [email protected] on Sat Sep 05, 2015 3:48 pm, edited 3 times in total.
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by Matt@VeritasPrep » Fri Sep 04, 2015 4:58 pm
[email protected] wrote:Hi tmontgomery,

What is the source of this question? I asked because the question is worded in an odd way, but I think the 'intent' of the question is the same as asking:

"There are 6 balls in a box - 2 red and 4 green. Balls will be randomly drawn one at a time and not replaced. What is the probability of selecting the 4 green balls on the first four draws?"

Under these conditions, we can deal with each individual 'event', then multiply the results.

Probability of drawing a green on the
first try: 4/6
second try: 3/5
third try: 2/4
fourth try: 1/3

(4/6)(3/5)(2/4)(1/3) = 1/15

Final Answer: B
The idea works, but it'd be more like this:

Prob(first letter is T, R, A, or M) = 4/7
Prob(second letter is T, R, or A) = 3/6
Prob(third letter is T or R) = 2/5
Prob(fourth letter is T) = 1/4

Which gives (4/7) * (3/6) * (2/5) * (1/4), or (1/35).
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by [email protected] » Sat Sep 05, 2015 9:29 am
Hi Matt,

Thanks for catching the typos; I've updated my solution accordingly.

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