Problem:
Pablo will decorate each of 6 identical white balls with either a striped or a dotted pattern, using either red or blue paint. He will decide on the color and pattern for each ball by flipping a fair coin for each of the 12 decisions he must make. After the paint dries, he will place the 6 balls in an urn. Frida will randomly select one ball from the urn and note its color and pattern. The events "the ball Frida selects is red" and "the ball Frida selects is striped" may or may not be independent, depending on the outcome of Pablo's coin flips. The probability that these two events are independent can be written as nm​, where m and n are relatively prime positive integers. What is m ? (Recall that two events A and B are independent if P(A and B)=P(A)⋅P(B).)
Answer Choices:
A. 243
B. 245
C. 247
D. 249
E. 251
Solution:
There are 46 ways to paint the balls, each equally likely. It remains to count the number of paintings for which the two given events are independent.
- If all the balls are red, then P( red )=1 and the events are independent regardless of P (striped). The same reasoning applies if all the balls are blue. This accounts for 2â‹…26=128 paintings.
- If 5 of the balls are red and s balls are striped, then
P( red )⋅P( striped )=65​⋅6s​
On the other hand P (red and striped) is one of the fractions 60​,61​,62​,63​,64​, or 65​, depending on how many red balls are striped. These are equal if and only if s=0 or s=6. There are 2⋅(56​)=12 ways for this to happen. A similar argument handles the case in which 5 balls are blue, giving 24 paintings in all.
- Suppose 4 balls are red. In order for the two given events to be independent, the fraction of red balls that are striped must equal the fraction of blue balls that are striped. This happens when all the balls are striped, or none of them are striped, or 2 of the red balls and 1 of the blue balls are striped. There are (24​)⋅(12​)=12 ways to choose the patterns in the last case, so this accounts for (46​)⋅(1+1+12)=210 situations. There are another 210 for which 2 balls are red.
- Suppose 3 balls are red. Again, in order for the two given events to be independent, the fraction of red balls that are striped must equal the fraction of blue balls that are striped. This happens when s red balls are striped and s blue balls are striped, where s∈{0,1,2,3}. The calculation in this case is
(36​)((03​)2+(13​)2+(23​)2+(33​)2)=20⋅20=400
In all there are 128+24+210+210+400=972 cases in which the color and the pattern of the drawn ball are independent, so the required probability is
212972​=1024243​
and the requested numerator is (A)243​.
The problems on this page are the property of the MAA's American Mathematics Competitions