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# There are six cards bearing numbers 2, 4, 5, 5, 5, 6. If two cards are

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Re: There are six cards bearing numbers 2, 4, 5, 5, 5, 6. If two cards are [#permalink]
This is how I solved it:

for positive difference to be 0 we will draw the cards like this:
(5,5)
for positive difference to be 1 we will draw the cards like this:
(6,5),(5,6),(5,4),(4,5)
for positive difference to be 2 we will draw the cards like this:
(2,4),(4,2),(6,4),(4,6)
for positive difference to be 3 we will draw the cards like this:
(5,2),(2,5)

==11 ways
Total ways of selecting any two cards:(5,5),(5,2),(2,5), (2,4),(4,2),(6,4),(4,6), (6,5),(5,6),(5,4),(4,5), (6,2), (2,6) ==13 ways

So the ans== 11/13

What did I do wrong? I literally thought that this is the only way we can pick cards without replacement. Also, since we are not replacing the cards, I thought we need to ensure that order matters.
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Re: There are six cards bearing numbers 2, 4, 5, 5, 5, 6. If two cards are [#permalink]
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beeblebrox wrote:
This is how I solved it:

for positive difference to be 0 we will draw the cards like this:
(5,5)
for positive difference to be 1 we will draw the cards like this:
(6,5),(5,6),(5,4),(4,5)
for positive difference to be 2 we will draw the cards like this:
(2,4),(4,2),(6,4),(4,6)
for positive difference to be 3 we will draw the cards like this:
(5,2),(2,5)

==11 ways
Total ways of selecting any two cards:(5,5),(5,2),(2,5), (2,4),(4,2),(6,4),(4,6), (6,5),(5,6),(5,4),(4,5), (6,2), (2,6) ==13 ways

So the ans== 11/13

What did I do wrong? I literally thought that this is the only way we can pick cards without replacement. Also, since we are not replacing the cards, I thought we need to ensure that order matters.

beeblebrox

I'll try to explain your mistake by first asking a question. If the deck instead included only 2, 4, 5, 5, 6 (missing one of the 5s), you would have ended up with the same result by deploying your method, right? Does having three 5s instead of two 5s matter? Of course it does! So your method can't be the right one. Why? When you listed (6,5), for example, you really needed to list (6,5a), (6,5b),(6,5c). The same is true for each time you listed a 5.

Okay, so how do we solve it?

How many total ways are there? We have 6 options for which card we draw first. We have 5 options for which card we draw second. So there are 6*5 = 30 total possible ordered pairs. Of those, the only ways to "lose" are (6,2) and (2,6). So 28/30 "win." That's 14/15.

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Re: There are six cards bearing numbers 2, 4, 5, 5, 5, 6. If two cards are [#permalink]
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Bunuel wrote:
There are six cards with numbers 2, 4, 5, 5, 5, and 6 on them. If two cards are randomly selected without replacement, what is the probability that the absolute difference between the numbers on these cards is at most 3?

A. $$\frac{8}{15}$$

B. $$\frac{9}{15}$$

C. $$\frac{10}{15}$$

D. $$\frac{12}{15}$$

E. $$\frac{14}{15}$$

M11-08

Official Solution:

There are six cards with numbers 2, 4, 5, 5, 5, and 6 on them. If two cards are randomly selected without replacement, what is the probability that the absolute difference between the numbers on these cards is at most 3?

A. $$\frac{8}{15}$$
B. $$\frac{9}{15}$$
C. $$\frac{10}{15}$$
D. $$\frac{12}{15}$$
E. $$\frac{14}{15}$$

The only pair with a difference greater than 3 is {2, 6}: $$(6 - 2 = 4) > 3$$. The probability of drawing this pair is $$\frac{1}{C^2_6}$$.

Thus, the probability that the absolute difference between the numbers on the cards is 3 or less is: $$P = 1 - \frac{1}{C^2_6}= \frac{14}{15}$$.

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Re: There are six cards bearing numbers 2, 4, 5, 5, 5, 6. If two cards are [#permalink]
Bunuel wrote:
Bunuel wrote:
There are six cards with numbers 2, 4, 5, 5, 5, and 6 on them. If two cards are randomly selected without replacement, what is the probability that the absolute difference between the numbers on these cards is at most 3?

A. $$\frac{8}{15}$$

B. $$\frac{9}{15}$$

C. $$\frac{10}{15}$$

D. $$\frac{12}{15}$$

E. $$\frac{14}{15}$$

M11-08

Official Solution:

There are six cards with numbers 2, 4, 5, 5, 5, and 6 on them. If two cards are randomly selected without replacement, what is the probability that the absolute difference between the numbers on these cards is at most 3?

A. $$\frac{8}{15}$$
B. $$\frac{9}{15}$$
C. $$\frac{10}{15}$$
D. $$\frac{12}{15}$$
E. $$\frac{14}{15}$$

The only pair with a difference greater than 3 is {2, 6}: $$(6 - 2 = 4) > 3$$. The probability of drawing this pair is $$\frac{1}{C^2_6}$$.

Thus, the probability that the absolute difference between the numbers on the cards is 3 or less is: $$P = 1 - \frac{1}{C^2_6}= \frac{14}{15}$$.

Hi Bunuel

Can you please let me know why are we taking $$C^2_6$$ when three numbers 5, 5, 5 repeat in the set.

For example if we were to find the number of ways one can select two digits from 2, 4, 5, 5, 5, and 6 would we use the following approach

1) Both the numbers don't repeat - Number of ways two numbers can be selected from 2, 4, 5, 5, 5, and 6 such that both the numbers do not repeat = $$C^2_4$$ = 6
2) The numbers repeat - Number of ways two numbers can be selected from 2, 4, 5, 5, 5, and 6 such that both the numbers repeat = $$1$$ = 1

Out of the seven selections made, only one selection in which both 2 and 6 were selected has absolute difference between the numbers on these cards > 3.

So shouldn't the probability = 5/7 ?

Please let me know what I am doing incorrect.
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Re: There are six cards bearing numbers 2, 4, 5, 5, 5, 6. If two cards are [#permalink]
2
Kudos
Obscurus wrote:
Bunuel wrote:
Bunuel wrote:
There are six cards with numbers 2, 4, 5, 5, 5, and 6 on them. If two cards are randomly selected without replacement, what is the probability that the absolute difference between the numbers on these cards is at most 3?

A. $$\frac{8}{15}$$

B. $$\frac{9}{15}$$

C. $$\frac{10}{15}$$

D. $$\frac{12}{15}$$

E. $$\frac{14}{15}$$

M11-08

Official Solution:

There are six cards with numbers 2, 4, 5, 5, 5, and 6 on them. If two cards are randomly selected without replacement, what is the probability that the absolute difference between the numbers on these cards is at most 3?

A. $$\frac{8}{15}$$
B. $$\frac{9}{15}$$
C. $$\frac{10}{15}$$
D. $$\frac{12}{15}$$
E. $$\frac{14}{15}$$

The only pair with a difference greater than 3 is {2, 6}: $$(6 - 2 = 4) > 3$$. The probability of drawing this pair is $$\frac{1}{C^2_6}$$.

Thus, the probability that the absolute difference between the numbers on the cards is 3 or less is: $$P = 1 - \frac{1}{C^2_6}= \frac{14}{15}$$.

Hi Bunuel

Can you please let me know why are we taking $$C^2_6$$ when three numbers 5, 5, 5 repeat in the set.

For example if we were to find the number of ways one can select two digits from 2, 4, 5, 5, 5, and 6 would we use the following approach

1) Both the numbers don't repeat - Number of ways two numbers can be selected from 2, 4, 5, 5, 5, and 6 such that both the numbers do not repeat = $$C^2_4$$ = 6
2) The numbers repeat - Number of ways two numbers can be selected from 2, 4, 5, 5, 5, and 6 such that both the numbers repeat = $$1$$ = 1

Out of the seven selections made, only one selection in which both 2 and 6 were selected has absolute difference between the numbers on these cards > 3.

So shouldn't the probability = 5/7 ?

Please let me know what I am doing incorrect.

We are selecting 2 cards out of a set of 6 cards, regardless of the numbers on them, the total number of ways to make this selection is 6C2. This formula represents the number of ways to choose 2 items from a set of 6 items, without considering the order of selection. In this context, the specific numbers on the cards are not relevant to the calculation. We are simply interested in the number of ways to choose any 2 cards from the given set of 6.

We can select the following pairs of cards:

$$2, 4\\ 2, 5_1\\ 2, 5_2\\ 2, 5_3\\ 2, 6 \\ 4, 5_1\\ 4, 5_2\\ 4, 5_3\\ 4, 6\\ 5_1, 5_2\\ 5_1, 5_3, \\ 5_1, 6\\ 5_2, 5_3\\ 5_2, 6\\ 5_3, 6$$
Re: There are six cards bearing numbers 2, 4, 5, 5, 5, 6. If two cards are [#permalink]
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