Problem Solving

If the 4 alphabets of TRAMPLER are picked, what is the probability that the picked up alphabets can be arranged to form TRAM?

Brent@GMATPrepNow wrote:

confuse mind wrote:If the 4 alphabets of TRAMPLER are picked, what is the probability that the picked up alphabets can be arranged to form TRAM?

Let's think of the 8 letters as: T, R1, A, M, P, L, E, and R2.

P(letters can spell TRAM) = (# of outcomes where letters can spell TRAM)/(# of ways to select 4 letters)

As always, begin with the denominator.
# of ways to select 4 letters
There 8 letters, and we must choose 4 of them.
Since order doesn't matter here, this can be accomplished in 8C4 ways.

Now the numerator.
# of outcomes where letters can spell TRAM
This can be accomplished in 2 ways:
1) T, R1, A, M
2) T, R2, A, M

So, P(letters can spell TRAM) = 2/8C4
= 2/70
= 1/35

Cheers,
Brent

If anyone is interested, we have a free video for calculating combinations (like 5C2) in your head. You can find it here: https://www.gmatprepnow.com/module/gmat-counting?id=789

I approached this problem differently as elaborated below:

Given word: 8 letters (R repeated; so 7 unique letters)

1st letter can be chosen in 7 ways; probability that this letter is (T,R,A or M) = 4/7
2nd letter can be chosen in 6 ways; probability that this letter is (T,R,A or M) = 3/6
3rd letter can be chosen in 5 ways; probability that this letter is (T,R,A or M) = 2/5
4th letter can be chosen in 4 ways; probability that this letter is (T,R,A or M) = 1/4

Therefore, total probability = (4/7) * (3/6) * (2/5) * (1/4) = 1/35

kartikshah wrote:

Brent@GMATPrepNow wrote:

confuse mind wrote:If the 4 alphabets of TRAMPLER are picked, what is the probability that the picked up alphabets can be arranged to form TRAM?

Let's think of the 8 letters as: T, R1, A, M, P, L, E, and R2.

P(letters can spell TRAM) = (# of outcomes where letters can spell TRAM)/(# of ways to select 4 letters)

As always, begin with the denominator.
# of ways to select 4 letters
There 8 letters, and we must choose 4 of them.
Since order doesn't matter here, this can be accomplished in 8C4 ways.

Now the numerator.
# of outcomes where letters can spell TRAM
This can be accomplished in 2 ways:
1) T, R1, A, M
2) T, R2, A, M

So, P(letters can spell TRAM) = 2/8C4
= 2/70
= 1/35

Cheers,
Brent

If anyone is interested, we have a free video for calculating combinations (like 5C2) in your head. You can find it here: https://www.gmatprepnow.com/module/gmat-counting?id=789

I approached this problem differently as elaborated below:

Given word: 8 letters (R repeated; so 7 unique letters)

1st letter can be chosen in 7 ways; probability that this letter is (T,R,A or M) = 4/7
2nd letter can be chosen in 6 ways; probability that this letter is (T,R,A or M) = 3/6
3rd letter can be chosen in 5 ways; probability that this letter is (T,R,A or M) = 2/5
4th letter can be chosen in 4 ways; probability that this letter is (T,R,A or M) = 1/4

Therefore, total probability = (4/7) * (3/6) * (2/5) * (1/4) = 1/35

I think you missed the R1, R2 case for denominator. No?