A box contains four bolts. Two of them, labeled #1 and#2, are 5mm in diameter, a

Question

A box contains four bolts. Two of them, labeled #1 and#2, are 5mm in diameter, and two of them, labeled #3 and #4 are 7mmin diameter. Bolts are randomly selected until a 5mm bolt isobtained. The outcomes are sequences that can be selected. Soone outcome is 1, and another is 342. Let X represent thenumber of bolts selected. Find the probabilty mass function of X. A box contains four bolts. Two of them, labeled #1 and#2, are 5mm in diameter, and two of them, labeled #3 and #4 are 7mmin diameter. Bolts are randomly selected until a 5mm bolt isobtained. The outcomes are sequences that can be selected. Soone outcome is 1, and another is 342. Let X represent thenumber of bolts selected. Find the probabilty mass function of X.

Explanation / Answer

Let random variable X be the number of bolts selected X can take possible values of 1, 2 or 3 P(X = 1) = 2/4=1/2 (The outcomes are 1 and 2) P(X = 2) = 4/(4x3) = 1/3 (The outcomes are 31, 32, 41, and 42)Total possible outcomes are 4x3 because there are 4 choices for thefirst pick and 3 remaining choices for the 2nd pick. P(X = 3) = 4/(4x3x2) = 1/6     (The outcomes are341, 342, 431 and 432) Total possible outcomes 4x3x2 since thereare 4 choices for the first pick, 3 remaining choices for the 2ndpick, and 2 remaining choices for the 3rd pick. Thus, we have the following prob mass function: X = 1, with prob = 1/2 X = 2, with prob = 1/3 X = 3, with prob = 1/6

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