rohit_gmat wrote:Source : GMAC's Test Prep software

Question (not word to word) : There are 5 apples in a basket. 1 of the 5 is rotten and the remaining 4 are good apples. If a person were to pick 2 apples simultaneously, what is the probability that he picks at least one bad apple?

Answer Choices : Not available...sorry

What's the best way to solve this?

Hi, when you pick objects simultaneously, it's the same as picking them one at a time, without replacement - so that's the easiest way to attack the question.

There are two outcomes that make us happy:

bad then good; and
good then bad.

Since we want EITHER outcome, we ADD the probabilities.

Bad then good = 1/5 * 4/4 = 4/20 = 1/5
Good then bad = 4/5 * 1/4 = 4/20 = 1/5

adding them together we get 2/5 as the final answer.

* * *

A second method would be to use the "1 minus" approach:

Prob(what you want) = 1 - prob(what you don't want)

Here, we don't want two good in a row, so:

Prob(good, good) = 4/5 * 3/4 = 12/20 = 3/5

prob(not good,good) = 1 - 3/5 = 2/5

* * *

A third method (and my personal favourite) is logic; if there's one bad apple in the bowl out of 5 apples in total, and if we're pulling x apples, there will always be an x/5 chance of getting the bad one:

1 pull, 1/5 shot;
2 pulls, 2/5 shot;
3 pulls, 3/5 shot;
4 pulls, 4/5 shot; and
5 pulls, sure thing!

Stuart Kovinsky wrote:
A third method (and my personal favourite) is logic; if there's one bad apple in the bowl out of 5 apples in total, and if we're pulling x apples, there will always be an x/5 chance of getting the bad one:

1 pull, 1/5 shot;
2 pulls, 2/5 shot;
3 pulls, 3/5 shot;
4 pulls, 4/5 shot; and
5 pulls, sure thing!

Thanks Stuart!
How would I use the "shot" method, if say there were 2 bad apples and 3 good ones, and we were asked to find the probability for picking at least 1 bad apple? Should I break it down - (one bad ... two bad) ?

rohit_gmat wrote:

Stuart Kovinsky wrote:
A third method (and my personal favourite) is logic; if there's one bad apple in the bowl out of 5 apples in total, and if we're pulling x apples, there will always be an x/5 chance of getting the bad one:

1 pull, 1/5 shot;
2 pulls, 2/5 shot;
3 pulls, 3/5 shot;
4 pulls, 4/5 shot; and
5 pulls, sure thing!

Thanks Stuart!
How would I use the "shot" method, if say there were 2 bad apples and 3 good ones, and we were asked to find the probability for picking at least 1 bad apple? Should I break it down - (one bad ... two bad) ?

You can't use the intuitive method unless it's a very simple situation (e.g. only 1 out of 5 is bad). For your second question, I'd use the "1 minus" approach. Assuming that we're still picking 2 apples, the only thing we don't want is both good, so:

Prob(at least 1 bad) = 1 - (prob both good)

Prob(at least 1 bad) = 1 - (3/5 * 2/5)

Prob(at least 1 bad) = 1 - 6/25 = 19/25

Since there is only 1 bad apple, the total no. of ways you can choose 1 bad and 1 good apple is 4.
You can do this by taking 1 bad apple (there is only 1 way it can be done) and 1 good apple out of 4 (it can be done 4 ways.
Multiply them: 1x4 = 4
Now you can pick 2 apples out of 5 in 5C2 ways, i.e. 10 ways. (total no. of different possible outcomes)
Probability: 4/10 = 2/5

srajan wrote:Since there is only 1 bad apple, the total no. of ways you can choose 1 bad and 1 good apple is 4.
You can do this by taking 1 bad apple (there is only 1 way it can be done) and 1 good apple out of 4 (it can be done 4 ways.
Multiply them: 1x4 = 4
Now you can pick 2 apples out of 5 in 5C2 ways, i.e. 10 ways. (total no. of different possible outcomes)
Probability: 4/10 = 2/5

Hi Srajan,

I am not so sure about that approach. If I were to use total # of desired outcomes / total # of possible outcomes, i would have smth like :

total # of desired outcomes : Good Apple + Bad Apple OR Bad Apple + Good Apple
Which is 2 (4!/3!) = 8

total # of possible outcomes : 5C2 = 5!/3! = 5 x 4 = 20

P = 8/20 = 2/5
....right?

rohit_gmat wrote:
total # of possible outcomes : 5C2 = 5!/3! = 5 x 4 = 20

Sorry I mean 5P2

rohit_gmat wrote:

rohit_gmat wrote:
total # of possible outcomes : 5C2 = 5!/3! = 5 x 4 = 20

Sorry I mean 5P2

Hi Rohit,

It will not be permutaion as it is a matter of selection. Order does not matter.
2 apples are selected together.
Therefore, it will be 5C2 not 5P2.
Also, the total no. of selections containing 1 bad apple will be 4, since you have to have 1 bad apple in that selection and the other apple can be selected by 4C1 ways.

If the order was important then the problem would have been handled differently.
The question would have been, 2 apples were selected one after another. What is the probability that at least one of them would be bad?
In this case, the total possible no. of ways of selections would be 5P2
The no. of ways you have at least 1 bad apple is 2x4P1
The answer would be same, but the steps would be different.

I hope it helps clear some doubt.