Statistics word problem

  • Thread starter war485
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  • #1
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Homework Statement



There is a group of four blood donors: A, B, C and D. Only A has type
A+ blood. Four blood samples, one from each individual, will be typed in random order until an A+ individual is identified. Let Y = {number of typings necessary to identify an A+ individual}. Compute the probability mass function.

Homework Equations



permutation (order matters): n! / (n - k)!

The Attempt at a Solution



let p = probability
p(Y=1) = 1/4
p(Y=2) = permute (1,1) * permute (1,3) / permute (2,4) = 1/4
p(Y=3) = permute (1,1) * permute (2,3) / permute (3,4) = 1/4
p(Y=4) = permute (1,1) * permute (3,3) / permute (4,4) = 1/4

Even though I think the specific order matters, it doesn't make sense that they each have a probability of 1/4. Did I miss something from the problem? Does this seem right/logical?
 

Answers and Replies

  • #2
Seems right to me.

[tex] P(Y=1) = \frac {1}{4} [/tex]

[tex] P(Y=2) = (\frac {3}{4}) (\frac {1}{3}) = \frac {3}{12} = \frac {1}{4} [/tex]

[tex] P(Y=3) = (\frac {3}{4}) (\frac {2}{3}) (\frac {1}{2}) = \frac {6}{24} = \frac {1}{4} [/tex]

[tex] P(Y=4) = (\frac {3}{4}) (\frac {2}{3}) (\frac {1}{2}) (\frac{1}{1})= \frac {6}{24} = \frac {1}{4} [/tex]
 

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