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505-555 Level|   Probability|            
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gmatt1476
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Answer: E

Basket 1:4 apples, 2 oranges. So total 6 fruits
Basket 2: 3 apples, 5 oranges. So total 8 fruits

Probability of selecting apple from basket 1 and orange from basket 2
= (4/6)*(5/8)
=5/12

Probability of selecting orange from basket 1 and apple from basket 2
=(2/6)*(3/8)
=1/8

Since order if fruit is not specified, we have to consider both these possibilities.

Total probability=(5/12)+(1/8) = 13/24

Posted from my mobile device
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gmatt1476
Carol purchased one basket of fruit consisting of 4 Apples and 2 Oranges and another basket of fruit consisting of 3 Apples and 5 Oranges. Carol is to select one piece of fruit at random from each of the two baskets. What is the probability that one of the two pieces of fruit selected will be an apple and the other will be an orange?

(A) 1/4
(B) 1/2
(C) 1/24
(D) 5/24
(E) 13/24

PS63210.02

question has asked ,what is the probability that one of the two pieces of fruit selected will be an apple and the other will be an orange

for first basket we take apple and orange from another ; 4/6 * 5/8 ; 5/12
2nd draw we take orange from 1st basket and apple from 2nd basket ; 2/6 * 3/8 ; 1/8
total ; 5/12+1/8 = 10+3/24 ; 13/24
OPTION E
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Basket 1

Apples----->4
Oranges -->2
Total------->6

Basket 2

Apples----->3
Oranges -->5
Total------->8

P(1 Apple from Basket 1 and 1 orange from Basket 2) = \(\frac{4}{6}*\frac{5}{8}\) = \(\frac{20}{48}\)

or

P(1 Orange from Basket 1 and 1 Apple from Basket 2) = \(\frac{2}{6}*\frac{3}{8}\) = \(\frac{6}{48}\)


Probability that one of the two pieces of fruit selected will be an apple and the other will be an orange = \(\frac{20}{48}\) + \(\frac{6}{48}\) = \(\frac{26}{48}\) = \(\frac{13}{24}\)

Therefore the answer is Option E

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Hello Brent,

Thanks for solving this.

While i understand your solution, i am a little confused about why this question doesn't require us to consider how many ways to choose say 1 Apple from 4 Apples in Basket A and 1 orange from the 2 oranges in basket A and so on.

I am yet to grasp when combinations come into probability questions and i would appreciate it if you could please shed more light on the combinations aspect.

Thank You

BrentGMATPrepNow
gmatt1476
Carol purchased one basket of fruit consisting of 4 Apples and 2 Oranges and another basket of fruit consisting of 3 Apples and 5 Oranges. Carol is to select one piece of fruit at random from each of the two baskets. What is the probability that one of the two pieces of fruit selected will be an apple and the other will be an orange?

(A) 1/4
(B) 1/2
(C) 1/24
(D) 5/24
(E) 13/24

PS63210.02

Let's name the baskets!
Basket A has 4 Apples and 2 Oranges
Basket B has 3 Apples and 5 Oranges


P(1 of each type) = P(select apple from basket A AND orange from basket B OR select orange from basket A AND apple from basket B)
= [P(apple from basket A) x P(orange from basket B)] + [P(orange from basket A) x P(apple from basket B)]
= [4/6 x 5/8] + [2/6 x 3/8]
= 20/48 + 6/48
= 26/48
= 13/24

Answer: E

Cheers,
Brent
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gmatt1476
Carol purchased one basket of fruit consisting of 4 Apples and 2 Oranges and another basket of fruit consisting of 3 Apples and 5 Oranges. Carol is to select one piece of fruit at random from each of the two baskets. What is the probability that one of the two pieces of fruit selected will be an apple and the other will be an orange?

(A) 1/4
(B) 1/2
(C) 1/24
(D) 5/24
(E) 13/24

PS63210.02

Solution:

The probability that an apple is selected from the first basket and an orange is selected from the second is 4/6 x 5/8 = 20/48. Similarly, the probability that an orange is selected from the first basket and an apple is selected from the second is 2/6 x 3/8 = 6/48. Therefore, the overall probability is 20/48 + 6/48 = 26/48 = 13/24.

Answer: E
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gmatt1476
Carol purchased one basket of fruit consisting of 4 Apples and 2 Oranges and another basket of fruit consisting of 3 Apples and 5 Oranges. Carol is to select one piece of fruit at random from each of the two baskets. What is the probability that one of the two pieces of fruit selected will be an apple and the other will be an orange?

(A) 1/4
(B) 1/2
(C) 1/24
(D) 5/24
(E) 13/24

PS63210.02

Solution:

The probability that an apple is selected from the first basket and an orange is selected from the second is 4/6 x 5/8 = 20/48. Similarly, the probability that an orange is selected from the first basket and an apple is selected from the second is 2/6 x 3/8 = 6/48. Therefore, the overall probability is 20/48 + 6/48 = 26/48 = 13/24.

Answer: E

PL. TELL THE COMBINATION APPROACH, THAT WILL BE GREAT HELP
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Using the combination approach

B1 - 4 A, 2 O; B2- 3 A, 5 O
Probability = No of fav outcomes/Total no of outcomes
Fav Outcomes = 4C1*5C1+2C1*3C1 (Selecting 1 A & 1O from B1 OR 1 O & 1 A from B2)
Total no of outcomes - 6C1*8C1 (Selecting 1 item from B1 AND Selecting one item from B2 - Question mentions that 1 item is picked from either basket at a time)

P=20+6/48=26/48=13/24
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4/6 * 5/8 = 20/48

2/6 * 3/8 = 6/48

total = 26/48 = 13/24
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SOLUTION:

A NEW OG 2021 QUESTION

Its based on the concepts of Counting and Probability

(Basket 1)-4 apples, 2 oranges => Total fruits=6 => Sample space =6

(Basket 2)-3 apples, 5 oranges=>Total fruits=8 => Sample space =8

The probability that one of the two pieces of fruit selected will be an apple and the other will be an orange =

Choosing Apple from first basket and orange from second + Choosing Orange from first and apples from second basket

=(4/6)*(5/8) + (2/6)*(3/8)
=5/12 + 1/8 =13/24 (OPTION E)

Hope this helps :thumbsup:
Devmitra Sen(Math)
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gmatt1476
Carol purchased one basket of fruit consisting of 4 Apples and 2 Oranges and another basket of fruit consisting of 3 Apples and 5 Oranges. Carol is to select one piece of fruit at random from each of the two baskets. What is the probability that one of the two pieces of fruit selected will be an apple and the other will be an orange?

(A) 1/4
(B) 1/2
(C) 1/24
(D) 5/24
(E) 13/24

PS63210.02

Solve the Official Questions more productively


Click here and solve 1000+ Official Questions with Video solutions as Timed Sectional Tests
and Dedicated Data Sufficiency (DS) Course

Answer: Option E

Video solution by GMATinsight


Get TOPICWISE: Concept Videos | Practice Qns 100+ | Official Qns 50+ | 100% Video solution CLICK.
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Hey everyone,

I've got a small doubt here -

After selecting one apple from basket 1 & orange from basket 2, why aren't we considering arranging them because the order of occurrence hasn't been mentioned here?
We may take the orange from basket 2 first or take apple first, don't we require to arrange them?

Thank you!
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Hi GMATters,

Here is my video solution to this question:

Best,

Rowan
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MartyTargetTestPrep, avigutman, can you please help in understanding why does the order (OA)/(AO) does not matter here? I have some understanding of it, but I haven't been able to articulate it.

Understand that we are looking at a fixed outcome, and hence it is different from P(Orange and Apple) from a set of 20 apples and oranges. In this latter case, we will have to use the arrangement factorial because we can pick orange in first try and apple in second or vice versa.

Please help me understand if I am thinking right.

Thank you!
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CrushTHYGMAT
MartyTargetTestPrep, avigutman, can you please help in understanding why does the order (OA)/(AO) does not matter here? I have some understanding of it, but I haven't been able to articulate it.

Understand that we are looking at a fixed outcome, and hence it is different from P(Orange and Apple) from a set of 20 apples and oranges. In this latter case, we will have to use the arrangement factorial because we can pick orange in first try and apple in second or vice versa.

Please help me understand if I am thinking right.

Thank you!
The easiest way to determine that order doesn't matter in this case is by noticing that we are selecting only one item from each basket. So, the selection of each item is completely independent of the selection of the other. Thus, the order in which the items are selected has absolutely no effect on the probability of selecting each item.
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MartyTargetTestPrep, thank you. My question is an extension to your advice that we shouldn't leave a question until and unless we precisely understand the reasoning as well as can articulate it.

Hope you have a great day!
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CrushTHYGMAT
MartyTargetTestPrep, thank you. My question is an extension to your advice that we shouldn't leave a question until and unless we precisely understand the reasoning as well as can articulate it.

Hope you have a great day!
Excellent! Taking that approach will serve you well in both quant and verbal. You have a great day too!
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Hi - I understand that by default we have assumed that all oranges are different form one another. (means choosing o1 will be a different case than choosing o2).
Was just wondering on the following 2 things -
1) In such types of questions, we consider the otheriwise similar objects (in the real world) to be different in EGMAT questions right? Like all the oranges are DISTINCT here? (if not explicitly mentioned in the question? )
2) In case the question mentioned that all oranges and apples are similar. The our answer would have been 2/4 i.e 1/2?

If someone could confirm on the above 2 points that would be great :)
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