Four nonnegative integers A,B,C and D, with 2A+B+C+D = 4, are chosen at random.
Find the probability that A+B+C+D = 3.
By the first equation A can be zero, 1 or 2.
If A is zero, the conditional probability is zero.
In this case B+C+D=4, which can happen in C(6,2) = 15 ways.
If A is 2, the conditional probability is zero.
In this case B,C and D are all zero. There is only 1 way for this.
If A is 1, B+C+D = 2 and the condition is met.
There are C(4,2) = 6 ways of this happening.
So there are 6 ways out of 22 that the second equation is true. That's 3/11 probability.
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Posted by Charlie
on 2016-09-23 10:44:58 |
solution