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
Probability of $5
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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
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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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.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
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?
Cheers,
Brent
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$2 coins=AJack 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?
$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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Perfect.pemdas wrote:$2 coins=AJack 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?
$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
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
Cheers,
Brent
- cypherskull
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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.
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.
Regards,
Sunit
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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
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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Let's put it more simplyBrent@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
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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Let's put it more simplyBrent@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
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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To begin, there's a problem with this statement: Prob. of retrieving 2 $2 coins = 2/4 = 1/2cypherskull 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.
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,
Brent
- cypherskull
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Thanks Brent! I see where I messed up now.
Regards,
Sunit
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Sunit
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Another approach is to list all of the possible outcomes for the 3 coins.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
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,
Brent