Problem Solving

Hey all,

Could figure this one out. A brief explanation would be much appreciated. Thanks!

A basket has 5 apples, one is spoiled. If Henry picked 2 apples simultaneously and at random, what is probability that the 2 selected apples will include the spoiled one.

a. 1�5
b. 2�10
c. 2�5
d. 1�2
e. 3�5

Hey PKW,
The probablity of choosing a good apple for the first selection is 4/5 since there are 4 good apples and one bad one. The probablity of choosing a good apple for the second selection is 3/4 since there are 3 good apples and one bad one left. Then you multiply these two and you get:
4/5 * 3/4 = 12/20 = 6/10 = 3/5

3/5 is the probablity that you will choose two good apples. So you can take 1 - 3/5 and you get: 2/5 which is the probablity that one of the selections will contain the spoiled apple.

Can someone show how to solve this problem using Combinations?

Thanks

funx wrote:Can someone show how to solve this problem using Combinations?

Thanks

Sure, although it's simpler with probability; in fact, even when we use combinations we use the probability formula.

Probability = # desired outcomes / total # of possibilities

Here, we're selecting 2 out of 5 apples, so the total # of possibilities is 5C2 = 5*4/2*1 = 10.

For # of desired outcomes, we want 1 bad apple out of 1 total bad apple and 1 good one out of 4 total good ones. So, we have:

1C1 * 4C1 = 1*4 = 4 desired outcomes.

Therefore, the probability of getting 1 bad apple on 2 draws is 4/10 = 2/5.

Brilliant, thank you so much Stuart!

Thanks guys. You all really make my life easier

pkw209 wrote:Hey all,

Could figure this one out. A brief explanation would be much appreciated. Thanks!

A basket has 5 apples, one is spoiled. If Henry picked 2 apples simultaneously and at random, what is probability that the 2 selected apples will include the spoiled one.

a. 1�5
b. 2�10
c. 2�5
d. 1�2
e. 3�5

Using my 5-step method for probability:
1) Lay out the number of events (here, we have 2 events):

_ _

2) Label each event with one specific example of the desired outcome:

_ _
S NS

(NOTE: S = spoiled, NS = not spoiled)

3) Label each event with its relevant probability and multiply across (here, we have a selection of elements where elements are removed), known as the "specific probability"

1/5 1 = 1/5
S NS

(NOTE: Since only non-spoiled remains for the second selection, the probability of selecting a non-spoiled is 100%)

4) Determine the number of ways in which the desired event can be presented (here, we have 2 ways):

S NS
NS S

5) Multiply the result of step 3 by the result of step 4:

1/5 x 2 = 2/5

(NOTE: When the specific probability is different for each possibility (which can occur), add the specific probabilities together).

(NOTE: I have noticed some comments about the accuracy of some math problems. I completely agree that GMAT students should be 100% that they have accurately re-created the question. Since the GMAT uses all the information in the questions, any mistake or omission can cause real problems. I once couldn't solve an exponent question that a student provided because the student listed the exponent as "8" instead of "18". Once it was changed to "18", the problem flowed easily).

Thanks Testrainer. I should have mentioned a disclaimer about pulling these questions directly from Zuleron's list of 198 questions, which were taken from GMAT PREP I. However, it appears that he or she abbreviated some of the question stems. That being said, they are official questions if that helps.

pkw209 wrote:Hey all,

Could figure this one out. A brief explanation would be much appreciated. Thanks!

A basket has 5 apples, one is spoiled. If Henry picked 2 apples simultaneously and at random, what is probability that the 2 selected apples will include the spoiled one.

a. 1�5
b. 2�10
c. 2�5
d. 1�2
e. 3�5

By combination:

4 good apples and 1 bad apple

You can only select one from each category to always have a bad apple. By fundamental counting principle

(4C1 x 1C1)/5C2 = 2/5

By Conditional Probability:
Let G be good apple and B be for bad apple.

You want want ( G AND B). or P( G AND B). Pick One apple first and see the outcome. There are two cases: First apple was G and first apple was B.

First Apple G:

P(G N B) = P(G) P(B|G)
( 4/5) ( 1/4= 1/5

First Apple was B: P(B) P(G|B)
(1/5)(4/4)=1/5

Adding both gives 2/5

Do you use the conditional probability as stated in dtweah's post when you are picking two objects simultaneously?

thanks guys!