Tres amigas, yet again!
A bag contains 10 red and 10 blue marbles.
- Ella takes a marble from the bag (without looking) and puts it in her pocket.
- Next, Bella takes a marble from the bag.
- Finally, Frabduzella takes a marble from the bag.
If Bella and Frabduzella both have blue marbles, what is the probability that Ella's marble is red?
If the probability is b a , where a and b are coprime positive integers, submit your answer as a + b .
Follow up to this problem
The answer is 14.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
2 solutions
Nice solution, same as mine :)
The main point of course is that the order is not important. But as in your problem, it requires the realization that information revealed "afterwards" (the fact that E is taller than B, or the fact that B & F got blue marbles) can affect the distribution of a random variable sampled "beforehand" (F's height relative to E or E's marble color.)
Curious to note that if the bag originally had N red and N + 2 blue marbles, and both Bella and Frabduzella took blue marbles, then Ella's marble was as likely red as blue, no matter what N ≥ 1 is.
There are other simpler solutions, but heres one through Baye's method:
P ( 2 B R ) = P ( 2 B ) P ( R Π 2 B ) , here R» Event that Ella got red, 2B» Event that the other 2 got blue.( Π = i n t e r s e c t i o n )
= ½ ( 1 9 1 0 ⋅ 1 8 9 ) + ½ ( 1 9 9 ⋅ 1 8 8 ) ½ ( 1 9 1 0 ⋅ 1 8 9 )
= 9 5
That's Bayes' Method, not Bayer's :)
The order with which they drew out the marbles is unimportant.
Bella and Frabduzella both took blue marbles.
That leaves 18 marbles that Ella could have chosen, 8 blues and 10 reds.
So, the probability she chose red is
1 8 1 0 = 9 5
5 + 9 = 1 4
Nice title, by the way! ;-)