Picking the 2 most expensive computers - Probability

gmattesttaker2 wrote:Hello,

Can you please assist with this:

A shop has 8 computers for sale; no two computers sell for the same price. If three
computers are picked at random, what is the probability of picking the two most
expensive ones?

OA: [spoiler]3/28[/spoiler]

Thanks a lot,
Sri

In DESCENDING order of price, let the 8 computers be A, B, C, D, E, F, G and H.

P(A and B are selected) = (good combinations)/(all possible combinations).

All possible combinations:
Number of ways to choose 3 computers from 8 = 8C3 = (8*7*6)/(3*2*1) = 56.

Good combinations:
ABC
ABD
ABE
ABF
ABG
ABH
Total options = 6.

P(A and B are selected) = 6/56 = [spoiler]3/28[/spoiler].

GMATGuruNY wrote:
In DESCENDING order of price, let the 8 computers be A, B, C, D, E, F, G and H.

P(A and B are selected) = (good combinations)/(all possible combinations).

All possible combinations:
Number of ways to choose 3 computers from 8 = 8C3 = (8*7*6)/(3*2*1) = 56.

Good combinations:
ABC
ABD
ABE
ABF
ABG
ABH
Total options = 6.

P(A and B are selected) = 6/56 = [spoiler]3/28[/spoiler].

Hi Mitch,
Could you please explain me where I am going wrong,
This is how I did.
let the 8 computers be A, B, C, D, E, F, G and H.
We need to choose A and B and any one.

So, P = 2/8 *1/7 *6/6 =1/28
I am getting,
P=1/28

What wrong I did Mitch?

Thanks in advance.

Regards,
Uva.

Uva@90 wrote: Hi Mitch,
Could you please explain me where I am going wrong,
This is how I did.
let the 8 computers be A, B, C, D, E, F, G and H.
We need to choose A and B and any one.

So, P = 2/8 *1/7 *6/6 =1/28
I am getting,
P=1/28

What wrong I did Mitch?

Thanks in advance.

Regards,
Uva.

Alternate approach:

Let E = either of the 2 most expensive computers and C = any of the 6 cheap computers.
We need to determine the probability both E's and 1 C are selected.

P(1st computer is an E) = 2/8. (Of the 8 computers, 2 are E's.)
P(2nd computer is an E) = 1/7. (Of the 7 remaining computers, 1 is an E.)
P(3rd computer is a C) = 6/6. (Of the 6 remaining computers, all 6 are C's.)
Since we want all 3 events to happen, we multiply the fractions:
2/8 * 1/7 * 6/6.

Since the C computer could be selected 1st, 2nd or 3rd -- for a TOTAL OF 3 WAYS to select 2 E's and 1 C --the product above must be multiplied by 3:
2/8 * 1/7 * 6/6 * 3 = 3/28.