Probability - intersection

Probability Level 1

A box contains balls of three colors, red, blue, and yellow. Three balls are randomly drawn from the box without replacement.

The table above shows the possible outcomes of each color. For example, $\red{\text{Red}}$ - $2 = 84$ means that there are $84$ outcomes in which two of the three drawn balls are red. And this is indicated by $\red{\text{Red}}$ - $3 = |A| = 4$ in the Venn diagram. Similarly, $\blue{\text{Blue}}$ - $3 = |B| = 20$ , and $\orange{\text{Yellow}}$ - $3 = |C| = 56$ .

Without finding the number of balls of each color, use the table to find the values of $D$ , $E$ , and $F$ . Submit the sum of digits of the values of $D$ , $E$ , and $F$ . For example, the sum of digits of $|A|$ , $|B|$ , and $|C|$ is $= \red 4 + \blue{2+0} + \orange{5+6} = 17$ .

43 45 25 40

1 solution

Chew-Seong Cheong
Aug 15, 2020

Note that $|A|+D+|C|$ indicates that there is no $\blue{\text{blue}}$ ball drawn. Therefore $|A|+D+|C| = \blue{\text {Blue}}$ - $0 = 220$ , $\implies D = 220 - |A|-|C| = 220-4-56 = 160$ . Similarly, $E = \orange{\text{Yellow}}$ - $0 - |A|-|B| = 120-4-20 = 96$ and $F = \red{\text{Red}}$ - $0 - |B|-|C| = 364 - 20 - 56 = 288$ .

Therefore the sum of digits $= \blue{1+6+0} + \orange{9+6} + \red{2+8+8} = \boxed{40}$ .

Probability - intersection

1 solution

0 pending reports