GMAT Club World Cup 2022 (DAY 7): Ronaldo puts an orange, an apple and

Answer is B

All possibility is 3*3*3*3=81
exact 2 type: 6+6+6+6+4+4+4=24+12=36
AAAO : 3*2*1*1 =6
BBBO : 3*2*1*1 =6
OOOA : 6
OOOB : 6
AAOO : 2*1*2*1=4
BBOO : 4
AABB : 4

Thus, probability that she got exactly 2 different types of fruit is 36/81 = 4/9 B

Bag 1: O, A, B
Bag 2: O, A, B
Bag 3: O,A, B
Bag 4: O,A, B

Now for 2 different fruits, combination could be: OO AA, OO BB, AA BB.

For any combination, Probability will be 1/9. So for 3 such combination, probability will be 3*1/9= 1/3

Option A is the correct answer

fruits are O,A,B and 4 bags so total fruits are 12
and P of drawing a fruit from 4 bags is 3^4 ; 81 and ways how these fruits can be drawn is 4c1
exactly 2 different types of fruit possibility
(o,o,o,a) ( o,o,o,b) ;(a,a,a,o) ( a,a,a,b) ; (b,b,b,o) ( b,b,b,a)
this can be arranged in4ways and total possibilities 24 as we have 6 such cases
(o,o,a,a) ( o,o,b,b)
(a,a,o,o) ( a,a,b,b)
(b,b,o,o) ( b,b,a,a)
4!/2!*2! ; 6 *3 ; 18
total 18+24 ; 42
42/81 ; 14/ 27


option D

Okay so this is how I did it:

The two fruits that are chosen by the 5th kid can be chosen in 3C2 ways = 3 ways.

From each lunchbox, the probability of picking a chosen fruit is 2/3.

Therefore, the total probability comes to be: 3*(2/3)*(2/3)*(2/3)*(2/3) = 16/27

However, this includes a case where the each of the chosen fruit is chosen from all 4 lunchboxes. So we have to deduct 2 from 16.

Answer is 14/27, i.e., option D

Exactly 2 fruits can be had in following Ways
AOOO - 4
AAOO - 6
AAAO - 4
ABBB - 4
AABB - 6
AAAB - 4
BOOO - 4
BBOO - 4
BBBO - 4

Required ways = 14 x 3
Total ways = 3 x 3 x 3 x 3

Prob = 14/27

Correct answer : Choice A

let Orange = O, Apple = A and Bannana = B

Number of ways all 3 fruits are the same = OOOO, AAAA, BBBB = 3 ways
Number of ways in which 3 different fruit + 1 fruit = OABA, OABB, OABO = 3 ways
Number of ways in which exactly 2 different sets of fruit are picked = OOAA, OOBB, AABB = 3 ways

So total number of ways = 3 + 3 + 3 = 9
probabilty = number of possible ways/ total number of ways
= 3 / 9
= 1 / 3
hence choice A is the correct answer

Total number of possible cases = (Comb (1of3))^4 =81

Possible cases of exactly two types of fruits are as follow : wether (Orange with apple) or (Orange with Banana) or (Apple with Banana)
to select one fruit (whatever the fruit is) from the 1st bag we have Comb (1 of 3) possibilities = 3
to select one fruit (whatever the fruit is) from the 2nd bag we have Comb (1 of 3) possibilities =3
Now while selecting from the 3rd bag, we'll have just one choice possible out of two so : Comb ( 1 of 2)=2
for the 4rth bag just one possible fruit is possible =1
so the possible cases are 3 x 3 x 2 x 1= 18
the probability then is : 18/81=1/3

Answer is A

Each box contains an orange, an apple and a banana.

We can either have

3 fruits of one kind and 1 fruit of another kind

OR

2 fruits of one kind and 2 fruits of another kind

3 fruits of one kind and 1 fruit of another kind

Probability of selecting any fruit from box 1 = 1
Probability of selecting from box 2 and box 3 the same fruit that was selected from box 1 = 1/3 * 1/3
Probability of selecting a different fruit from box 4 that that was selected from box 1 = 2/3

This can be done in 3!/2! = 3 ways

Net probability = \(1 * \frac{1}{9} * \frac{2}{3} * 3 = \frac{2}{9}\)

2 fruits of one kind and 2 fruits of another kind

Probability of selecting any fruit from box 1 = 1
Probability of selecting from box 2 the same fruit that was selected from box 1 = 1/3
Probability of selecting a different fruit than that was selected in box 1 = 2/3
Probability of selecting from box 4 the same fruit that was selected from box 2 = 1/3

This can be done in 3!/2! = 3 ways

Net probability = \(1 * \frac{1}{9} * \frac{2}{3} * 3 = \frac{2}{9}\)

Total probability = \(\frac{2}{9} + \frac{2}{9 }= \frac{4}{9}\)

IMO B

Ronaldo puts an orange, an apple and a banana in each of his 4 children's lunch boxes. The 5th child runs into the kitchen and randomly grabs one fruit from each of the 4 bags. What is the probability that she got exactly 2 different types of fruit ?

(A) 1/3
(B) 4/9
(C) 13/27
(D) 14/27
(E) 16/27


probability = 6*4C2*1/81 = 6*6*1/81 = 4/9

answer B

[quote="Bunuel"]Ronaldo puts an orange, an apple and a banana in each of his 4 children's lunch boxes. The 5th child runs into the kitchen and randomly grabs one fruit from each of the 4 bags. What is the probability that she got exactly 2 different types of fruit ?

(A) 1/3
(B) 4/9
(C) 13/27
(D) 14/27
(E) 16/27
Number of fruits in each bag=[Apple,Orange,banana]=3
Total number of ways the 5th child can select 4 fruits randomly out of the 4 bags =3*3*3*3=81
Now total number of ways by which the child will select exactly 2 different type of fruits=If the child picks 2 fruits of each type + If the child picks 3 fruits of 1 type and 1 of the other type
O means Orange
A means Apple
B means banana]
Possible combination
OAAA - Now total number of arrangements for this combination =4!/3!=4(These different arrangements are for picking individual fruits from each basket)
OBBB-total number of arrangements for this combination =4!/3!=4
OOAA-total number of arrangements for this combination =4!/(2!*2!)=6
OOBB-total number of arrangements for this combination =4!/(2!*2!)=6
OOOA-total number of arrangements for this combination =4!/(3!*1!)=4
OOOB-total number of arrangements for this combination =4!/(3!*1!)=4
ABBB-total number of arrangements for this combination =4!/(3!*1!)=4
AAAB-total number of arrangements for this combination =4!/(3!*1!)=4
AABB-total number of arrangements for this combination =4!/(2!*2!)=6
Thus total number of case =4+4+6+6+4+4+4+4+6
=42
Thus probability=42/81
=14/27
Answer -D

Total number of possibilities by which the 5th child can select one random fruit from each bag = (3C1)^4=81

Now let's consider the scenarios in which the child will pick exactly two types of fruits. 3 kinds of fruits are Orange (O), Apple (A) and Banana(B).

1 2 3 4
O A A A -> 4!/3! = 4
O B B B -> 4!/3! = 4
O O A A -> 4!(2!*2!) = 6
O O B B -> 4!(2!*2!) = 6
O O O A -> 4!/3! = 4
O O O B -> 4!/3! = 4
A B B B -> 4!/3! = 4
A A B B -> 4!(2!*2!) = 6
A A A B -> 4!/3! = 4
Total scenarios=42

Hence probability that she got exactly 2 different types of fruit
=42/81 = 14/27

ANS: D

Imo B

Lets say the 5th child makes this selection: Apple Apple Banana Orange
Another selection can be Apple Banana Apple Orange
Number of ways = 4!/2!
Similarly he or she can also choose 2 oranges and rest two different fruit or 2 bananas and rest two different.
Hence Total number of favorable ways = (4!/2!) * 3 = 36

Total cases = 3 ^ 4

Probability = Favorable / Total = 36 / 3^4 = 4/ 9

we are required to find the probability that the 5th child got exactly 2 different types of fruit ?
this can be OOAA or AAAB and similar combinations
which is = 3*(5) + 6*(4) = 39
No of ways of selecting 1 fruit from each of the four boxes containing 3 fruits is = 3C1 *3C1*3C1*3C1 = 81
therefore required prob = 39/81 = 13/27
Answer:C

Let orange - O
Let apple - A
Let banana - B

All 4 lunch boxes have 1A + 1B + 1O

Total number of ways 1 fruit can be grabbed from each 4 boxes: [3C1]^4 = 81 ways
Total number of ways exactly 2 different fruits = 14x 3 = 42
(P) = 42/81 = 14/27 [Option D]

Number of ways to select a fruit from each lunchbox:

3^4 = 81

Number of ways to select the same fruit from each lunchbox:

3

Number of ways to select 3 different fruits:

3 ways to pick which fruit is repeated * 4!/2 ways they are selected from the lunchboxes

= 36


Number of ways two different fruits can be selected:

81-3-36 = 42

Probability: 42/81 = 14/27

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Answer: B

Ronaldo puts an orange, an apple and a banana in each of his 4 children's lunch boxes. The 5th child runs into the kitchen and randomly grabs one fruit from each of the 4 bags. What is the probability that she got exactly 2 different types of fruit ?

Total no. of ways 5th child can grab a fruit = 3*3*3*3 = 81
No. of ways to pick two oranges and two apples = 4
No. of ways to pick two oranges and two banana = 4
No. of ways to pick two apples and two banana = 4

No. of ways to pick 3 oranges and a fruit other than orange = 2*4 = 8
No. of ways to pick 3 apples and a fruit other than apple = 2*4 = 8
No. of ways to pick 3 banana and a fruit other than banana = 2*4 = 8

Therefore, Required probability = (12+24) / 81
= 36/81 = 12/27 = 4/9

If all fruits were the same -> 1 * 3(one for each fruit) = 3
If 3 were different (last will have to be repeated) = 1*3 = 3
If exactly 2 different types of fruit were chosen -> AAOO AABB OOBB -> 3

probability that she got exactly 2 different types of fruit = 3/9 = 1/3

ANS - A