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Probability of $5

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by Brent@GMATPrepNow » Thu Aug 23, 2012 6:36 am
Given the ambiguity of the probability question at https://www.beatthegmat.com/probability-t118199.html , I decided to rewrite the question to be a little more GMAT-like.

Jack has two $2 coins and two $1 coins in his pocket. If he randomly selects 3 coins without replacement, then what is the probability that he will extract exactly $5?
A) 1/8
B) 1/4
C) 3/8
D) 1/2
E) 3/4

Answer = D

Challenge: Let's see how many different approaches are possible here. Off the top of my head, I think there are at least 3 or 4 ways to solve this.

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by rijul007 » Thu Aug 23, 2012 1:15 pm
If he takes out 3 coins out of his pocket, he would have exactly two coins of a kind. There would either be two $1 coins or two $2 coins.

Two $1 coins and one $2 coin = $4
Two $2 coins and one $1 coin = $5

There is one favorable outcome out of the two possibilites, hence the probability would be 1/2.

Option D
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by Brent@GMATPrepNow » Thu Aug 23, 2012 3:36 pm
rijul007 wrote:If he takes out 3 coins out of his pocket, he would have exactly two coins of a kind. There would either be two $1 coins or two $2 coins.

Two $1 coins and one $2 coin = $4
Two $2 coins and one $1 coin = $5

There is one favorable outcome out of the two possibilites, hence the probability would be 1/2.

Option D
That's a good approach, but if we're going to go that route, we need to show that the two outcomes are equally likely. That step is crucial.

To demonstrate this, consider this example question: What is the probability of being struck by lightning today? Well, we could say that there are 2 outcomes: a) getting struck by lightning and b) not getting struck by lightening. So, one might conclude that the probability of getting struck by lightening is 1/2. This, of course, is not the case since the 2 outcomes are not equally likely.

Now, in the original question, the $4 outcome is just as likely as the $5 outcome, so the probability is, indeed, 1/2.

Okay, so that's one approach. Any others?

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by pemdas » Thu Aug 23, 2012 5:11 pm
Jack has two $2 coins and two $1 coins in his pocket. If he randomly selects 3 coins without replacement, then what is the probability that he will extract exactly $5?
$2 coins=A
$1 coins=B

total selections of 3 coins=4C3 (value 4)
good selections($5)--> 2A out of 2A and 1B out of 2B, 2C1*2C2=2 (value 2)
probability=2/4=1/2
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by Brent@GMATPrepNow » Thu Aug 23, 2012 8:56 pm
pemdas wrote:
Jack has two $2 coins and two $1 coins in his pocket. If he randomly selects 3 coins without replacement, then what is the probability that he will extract exactly $5?
$2 coins=A
$1 coins=B

total selections of 3 coins=4C3 (value 4)
good selections($5)--> 2A out of 2A and 1B out of 2B, 2C1*2C2=2 (value 2)
probability=2/4=1/2
Perfect.

Here's another solution:

Rather than select 3 coins to withdraw, select one coin to remain in the pocket (it's the same thing)

Question: What coin must remain in order to withdraw $5?
Answer: The $1 coin must remain.

What's the possibility of selecting a $1 coin to remain?
Well, there are 4 coins altogether, and 2 of them are $1 coins.
So, the probability is 2/4 (or 1/2) that a $1 coin is selected to remain.

The answer is still D

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by cypherskull » Thu Aug 23, 2012 11:01 pm
I took the following approach and I'm getting 1/4 as the answer. Please tell me what am I doing wrong.

Soln:
The only way we could retrieve $5 in three coins is if 2 are $2 and 1 is $1.

Prob. of retrieving 2 $2 coins = 2/4 = 1/2
Prob. of retrieving 1 $1 coin (after the above 2 coins have been taken out) = 1/2

Total probability = (1/2)*(1/2) = 1/4.
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by mssharsha » Fri Aug 24, 2012 12:24 am
3 coins from 4 coins in this case shoud contain either 2 2$ coins or 2 1$ coins.. of them only 1 combination makes the sum 5$.. ie its probability is one in 2 cases. i.e 1/2
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by niketdoshi123 » Fri Aug 24, 2012 3:47 am
On method could be counting the number of ways of selecting the three coins.

S => selected
n => not selected

$2_$2__$1_$1__Total
S__ S__ S__ n__ = $5
S__ S__ n__ S__ = $5
S__ n__ S__ S__ = $4
n__ S__ S__ S__ = $4

Total # of ways = 4
# of ways he will exactly extract $5 = 2
P($5) = 2/4 = 1/2
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by SmartAssJun » Fri Aug 24, 2012 4:47 am
Brent@GMATPrepNow wrote:Given the ambiguity of the probability question at https://www.beatthegmat.com/probability-t118199.html , I decided to rewrite the question to be a little more GMAT-like.

Jack has two $2 coins and two $1 coins in his pocket. If he randomly selects 3 coins without replacement, then what is the probability that he will extract exactly $5?
A) 1/8
B) 1/4
C) 3/8
D) 1/2
E) 3/4

Answer = D

Challenge: Let's see how many different approaches are possible here. Off the top of my head, I think there are at least 3 or 4 ways to solve this.

Cheers,
Brent
Let's put it more simply
If you leave either of the two $1 coins in the pocket, then you'll get $5
and there's 2 $1 coins you have, which is half of them.
So the probably is 1/2
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by SmartAssJun » Fri Aug 24, 2012 4:47 am
Brent@GMATPrepNow wrote:Given the ambiguity of the probability question at https://www.beatthegmat.com/probability-t118199.html , I decided to rewrite the question to be a little more GMAT-like.

Jack has two $2 coins and two $1 coins in his pocket. If he randomly selects 3 coins without replacement, then what is the probability that he will extract exactly $5?
A) 1/8
B) 1/4
C) 3/8
D) 1/2
E) 3/4

Answer = D

Challenge: Let's see how many different approaches are possible here. Off the top of my head, I think there are at least 3 or 4 ways to solve this.

Cheers,
Brent
Let's put it more simply
If you leave either of the two $1 coins in the pocket, then you'll get $5
and there's 2 $1 coins you have, which is half of them.
So the probably is 1/2
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by Brent@GMATPrepNow » Fri Aug 24, 2012 7:06 am
cypherskull wrote:I took the following approach and I'm getting 1/4 as the answer. Please tell me what am I doing wrong.

Soln:
The only way we could retrieve $5 in three coins is if 2 are $2 and 1 is $1.

Prob. of retrieving 2 $2 coins = 2/4 = 1/2
Prob. of retrieving 1 $1 coin (after the above 2 coins have been taken out) = 1/2

Total probability = (1/2)*(1/2) = 1/4.
To begin, there's a problem with this statement: Prob. of retrieving 2 $2 coins = 2/4 = 1/2

I can see why you are saying it (there are 2 two-dollar coins out of 4), but this doesn't mean the probability of selecting those two coins is 2/4. This solution doesn't take into account how many coins we are selecting. For example, if we were to select 4 coins, the probability would be 1 that both two-dollar coins are selected. Similarly, if we were to select 1 coin, the probability would be 0 that both two-dollar coins are selected. So, where does the fact that we are drawing 3 coins fit into this equation?

Now, you've correctly identified that the only way to retrieve $5 is to have two $2 coins and one $1 coin.

At this point, we should ask, "What needs to happen for this event to occur?"
We get: P($5 in total) = P(1st coin is $2, 2nd coin is $2, 3rd coin is $1 OR 1st coin is $2, 2nd coin is $1, 3rd coin is $2 OR 1st coin is $1, 2nd coin is $2, 3rd coin is $2)
= P(1st coin is $2, 2nd coin is $2, 3rd coin is $1) + P(1st coin is $2, 2nd coin is $1, 3rd coin is $2) + P(1st coin is $1, 2nd coin is $2, 3rd coin is $2)
= (2/4 x 1/3 x 2/2) + (2/4 x 2/3 x 1/2) + (2/4 x 2/3 x 1/2)
= (1/6) + (1/6) + 1/6
= 1/2
= D

Cheers,
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by cypherskull » Fri Aug 24, 2012 10:00 am
Thanks Brent! I see where I messed up now.
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by Brent@GMATPrepNow » Sat Aug 25, 2012 12:22 pm
Brent@GMATPrepNow wrote: Jack has two $2 coins and two $1 coins in his pocket. If he randomly selects 3 coins without replacement, then what is the probability that he will extract exactly $5?
A) 1/8
B) 1/4
C) 3/8
D) 1/2
E) 3/4

Answer = D
Another approach is to list all of the possible outcomes for the 3 coins.
To do this, let's distinguish the four coins as follows: 1A, 1B, 2A and 2B

So, the possible outcomes when we withdraw 3 coins are:
1A, 1B, 2A
1A, 1B, 2B
1B, 1A, 2A
1B, 1A, 2B
1A, 2A, 1B
1A, 2B, 1B
1B, 2A, 1A
1B, 2B, 1A
2A, 1A, 1B
2A, 1B, 1A
2B, 1A, 1B
2B, 1B, 1A
2A, 2B, 1A
2A, 2B, 1B
2B, 2A, 1A
2B, 2A, 1B
2A, 1A, 2B
2A, 1B, 2B
2B, 1A, 2A
2B, 1B, 2A
1A, 2A, 2B
1B, 2A, 2B
1A, 2B, 2A
1B, 2B, 2A

Of the 24 equally-likely possible outcomes, 12 yield a sum of $5.

So, the probability is 12/24 ([spoiler]= 1/2 = D[/spoiler])

Cheers,
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