If a card is randomly selected, what is the probability that the card is any one suit, e.g., a diamond, heart, spade, or club?
1 answer:
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Answer:
a) h(x; 7, 5, 12) = (⁵Cₓ)( ⁷C₇₋ₓ) / (¹²C₇)
b) 2.92
Step-by-step explanation:
a)
Here
Number of interviewees = N = 12
Number of job openings = M = 5
Interviews schedules for the first day = n = 7
N − M = 12 - 5 = 7
Using hypergeometric distribution:
Let X be the no. top four candidates interviewed on first day.
The probability mass function of X:
P(X = x) =
/
It can be written as:
h(x; n, M, N) =
/
= (5Cx) (7C7-x) / (12C7)
= (⁵Cₓ)( ⁷C₇₋ₓ) / (¹²C₇)
h(x; 7, 5, 12) = (⁵Cₓ)( ⁷C₇₋ₓ) / (¹²C₇)
b)
The expectation is: E(X) = np
E(X) = n * M/N
= 7 * 5/12
= 7 * 0.41667
= 2.9167