A certain basket contains 10 apples - Probability question

[rrampriya]

A certain basket contains 10 apples, 7 of which are red and 3 are green. If 3 different apples are to be selected at random from the basket, what is the probability that 2 of the apples selected will be red and 1 will be green?

a) 7/40
b) 7/20
c) 49/100
d) 21/40
e) 7/10

Can someone help to solve this problem?

[gokul_nair1984]

A certain basket contains 10 apples, 7 of which are red and 3 are green. If 3 different apples are to be selected at random from the basket, what is the probability that 2 of the apples selected will be red and 1 will be green?

7 Red+ 3 Green=10 Apples

Probability that 2 of the apples selected will be red means selecting 2 red apples from 7 red ones. This can be mathematically stated as 7C2 or C(7,2)------(1)

Similarly, Probability of selecting 1 green apple from 3 green ones can be mathematically stated as 3C1 or C(3,1)----(2)

In Totality you are selecting 3 apples from a group of 10 which can be stated as 10C3 or C(10,3)----(3)

So the combined probablity can be expressed as
[C(7,2)*C(3,1)]/C(10,3) and this simplifies to 21/40

[RonPurewal]

A certain basket contains 10 apples, 7 of which are red and 3 are green. If 3 different apples are to be selected at random from the basket, what is the probability that 2 of the apples selected will be red and 1 will be green?

7 Red+ 3 Green=10 Apples

Probability that 2 of the apples selected will be red means selecting 2 red apples from 7 red ones. This can be mathematically stated as 7C2 or C(7,2)------(1)

Similarly, Probability of selecting 1 green apple from 3 green ones can be mathematically stated as 3C1 or C(3,1)----(2)

In Totality you are selecting 3 apples from a group of 10 which can be stated as 10C3 or C(10,3)----(3)

So the combined probablity can be expressed as
[C(7,2)*C(3,1)]/C(10,3) and this simplifies to 21/40

nicely done.

[aaron.1choi]

I did it a different way.

First: calculate all possible ways to choose 3 apples from 10.
10x9x8 = 720

Second: calculate the number of ways 2 green and 1 red ball can be chosen.

(ways to choose 1st green ball) x (ways to choose 2nd green ball) x (ways to choose 1st red ball)
7C2 x 6C1 x 3C1

= (7!)/(2!5!)x(6!)/(1!5!)x(3!)/(1!2!) = ((7x6)/2)x6x3 = 378

probability of 2G1R combination = number ways for 2G1R/total ways
=378/720
=21/40

[RonPurewal]

that works too.

is this problem actually from the GMAT PREP software?

[dudewheysmehemail]

that works too.

is this problem actually from the GMAT PREP software?

Hi Ron,
Yes this question is from the GMAT Prep test software. I took Practice Test Two today and I got it. Unfortunately my answer was wrong :-(

I was running short of time and this came up as #34, so I tried to rush through it and made an educated guess. I tried:
2 red apples and 1 green apple =
7/10 * 6/9 * 3/8 = 7/40.

I understand the other approaches listed above, but in this approach, I guess I didn't account for the 3 ways that this can happen (7/40 * 3 = 21/40).

Is this the right logic?

Many thanks!
--Rishi

[mithunsam]

that works too.

is this problem actually from the GMAT PREP software?

Hi Ron,
Yes this question is from the GMAT Prep test software. I took Practice Test Two today and I got it. Unfortunately my answer was wrong :-(

I was running short of time and this came up as #34, so I tried to rush through it and made an educated guess. I tried:
2 red apples and 1 green apple =
7/10 * 6/9 * 3/8 = 7/40.

I understand the other approaches listed above, but in this approach, I guess I didn't account for the 3 ways that this can happen (7/40 * 3 = 21/40).

Is this the right logic?

Many thanks!
--Rishi

That is correct.

7/10 * 6/9 * 3/8 gives us the scenario of picking two red balls first, then a green ball (RRG). Other possibilities are GRR and RGR. Therefore, we need to multiply with 3.

[RonPurewal]

7/10 * 6/9 * 3/8 gives us the scenario of picking two red balls first, then a green ball (RRG). Other possibilities are GRR and RGR. Therefore, we need to multiply with 3.

yep