Project 44: Tree Prob B

The probability of red then red is (2/5)(1/3) = 2/15

The probability of red then blue is (2/5)(2/3) = 4/15

The probability of blue then red is (3/5)(4/9) = 12/45 = 4/15

The probability of blue then blue is 1(3/5)(5/9) = 5/45 = 5/15  (or 1/3)

The sum of these four probabilities is 15/15 = 1

Project 44: Variables 'pA', 'pBGivenA', and 'pNBGivenNA' have been initialized.  

the probability of A (2/5 in the first example)

 the probability of B given A (1/3 in the first example)

the probability of not B given not A (5/9 in the first example)

Task: Appropriately initialize the value of 'pB' that represents the probability of B.  The value of p(B) is 9/15 (4/15 + 5/15) for the first example which would represent the probability that the second marble was blue.

**If your code works for 5 test cases, you can enter your e-mail address

Universal Computational Math Methods:

pow(5,2) returns 25.0

abs(-3.0) returns 3

sqrt(49.0) returns 7.0