Highly Probable!!!

Probability Level 5

You have 3 boxes in front of you.

First box has 3 yellow \color{#CEBB00}{3 \text{ yellow}} , 4 green \color{#20A900}{4 \text{ green}} and 5 blue \color{#3D99F6}{5 \text{ blue}} balls.

Second box has 8 red \color{#D61F06}{8 \text{ red}} , 1 green \color{#20A900}{1 \text{ green}} and 3 blue \color{#3D99F6}{3 \text{ blue}} balls.

Third box has 6 red \color{#D61F06}{6 \text{ red}} , 4 yellow \color{#CEBB00}{4 \text{ yellow}} and 2 green \color{#20A900}{2 \text{ green}} balls.

Now, you randomly select one out of 3 boxes and select a ball from it randomly, note it's colour and put it back into the box. You do this 3 times .

Now, because of some sort of magic, the probability of picking first box is 1 2 \frac{1}{2} , second box is 1 6 \frac{1}{6} and third box is 1 3 \frac{1}{3} . Probability of picking any ball is same once you select a box.

What is the probability that you get all the three balls of the same colour. Your answer can be written as x 7 2 3 \displaystyle \frac{x}{72^3} where x x is a positive integer. Enter your answer as x x .

Try my probability set

All of my problems are original

Difficulty: † † † † † \dagger \dagger \dagger \dagger \dagger


The answer is 23658.

Aryan Sanghi by Aryan Sanghi , 17, India

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1 solution

Aryan Sanghi
Jun 2, 2020

Let's start by considering probability of selecting each coloured ball individually

Red Ball \color{#D61F06}{\text{Red Ball}}

P R = 1 6 × 8 12 + 1 3 × 6 12 \color{#D61F06}{P_R} = \frac{1}{6} × \frac{8}{12} + \frac{1}{3} × \frac{6}{12}

P R = 20 72 \color{#D61F06}{P_R} = \frac{20}{72}

Blue Ball \color{#3D99F6}{\text{Blue Ball}}

P B = 1 2 × 5 12 + 1 6 × 3 12 \color{#3D99F6}{P_B} = \frac{1}{2} × \frac{5}{12} + \frac{1}{6} × \frac{3}{12}

P B = 18 72 \color{#3D99F6}{P_B} = \frac{18}{72}

Yellow Ball \color{#CEBB00}{\text{Yellow Ball}}

P Y = 1 2 × 3 12 + 1 3 × 4 12 \color{#CEBB00}{P_Y} = \frac{1}{2} × \frac{3}{12} + \frac{1}{3} × \frac{4}{12}

P Y = 17 72 \color{#CEBB00}{P_Y} = \frac{17}{72}

Green Ball \color{#20A900}{\text{Green Ball}}

P G = 1 2 × 4 12 + 1 6 × 1 12 + 1 3 × 2 12 \color{#20A900}{P_G} = \frac{1}{2} × \frac{4}{12} + \frac{1}{6} × \frac{1}{12} + \frac{1}{3} × \frac{2}{12}

P G = 17 72 \color{#20A900}{P_G} = \frac{17}{72}

Probability

P = P R 3 + P B 3 + P Y 3 + P G 3 P = \color{#D61F06}{P_R^3} \color{#333333}{+} \color{#3D99F6}{P_B^3} \color{#333333}{+} \color{#CEBB00}{P_Y^3} \color{#333333}{+} \color{#20A900}{P_G^3}

P = 23658 7 2 3 \color{#3D99F6}{\boxed{\color{#333333}{P = \frac{23658}{72^3}}}}

4 Helpful 1 Interesting 4 Brilliant 1 Confused

Great solution! I feel though that it would be useful to mention that each try is independent, and so that's why we're allowed to compute each probability separately and then multiply them at the end.

alazrabed . - 9 months, 4 weeks ago

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