A box contains 4 chocobars and 4 ice creams. Tom eats 3 of them, by ra

Question

A box contains 4 chocobars and 4 ice creams. Tom eats 3 of them, by randomly choosing. What is the probability of choosing 2 chocobars and 1 icecream?

(A) \frac{2}{7}

(B) \frac{3}{7}

(C) \frac{2}{3}

(D) \frac{1}{7}

(E) \frac{1}{3}

GMATinsight · Answer

Ways of choosing 2 chocobars and 1 icecream out of 4 chocobars and 4 icecreams = 4C2*4C1 = 6*4 = 24

Total ways of choosing any 3 items out of 8 items (chocobars+Icecreams) = 8C3 = 56

Required probability = 24/56 = 3/7

Answer: Option B

karan12345   Bunuel  Please correct the OA it shoul dbe Option B instead of Option D

karan12345 · Answer

Thanks for correcting me, I have updated the OA.

GunjanMadan1 · Answer

I think Answer should be \frac{1}{7} only as order does not matter here. Any of the Chocobar picked from 4 is the same.

Hence

Probability of choosing 1 chocobar = 4/8 = 1/2
After taking out 1 chocobar, the total number is 7.
Probability of choosing 2nd chocobar = 3/7
Probability of choosing 1 icecream out of a total of 6 = 4/6 = 2/3

So the final probability of choosing 2 chocobars and 1 icecream = 1/2 * 3/7 * 2/3 = 1/7

Now, no where in question is mentioned the order whether chocobar is picked first or ice cream. As if they are picked together. So {C,C,I}, {i,C,C} or {C,I,C} all are same here and should not be counted thrice.

A box contains 4 chocobars and 4 ice creams. Tom eats 3 of them, by ra

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