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rgrocks
11-04-2009, 03:40 PM
a bin of 5 transistors is known to have 3 defective ones. The transistors are to be tested one at a time until the defective ones are identified. denoted by N1, the number of test made to determine the first defective transistor and N2 the number of additional test until the second defective transistor is found. find the joint probability mass function of N1 and N2?

i have 0 idea where to even begin this.

chrisr
12-24-2009, 09:39 PM
There are 6 possibilities for this golden oldie,
sorry this one's late!

N[sub:2oia1avp]1[/sub:2oia1avp] = 1
N[sub:2oia1avp]2[/sub:2oia1avp] = 1

P = \frac{3}{5}\frac{2}{4} = \frac{3}{10}

N[sub:2oia1avp]1[/sub:2oia1avp] = 1
N[sub:2oia1avp]2[/sub:2oia1avp] = 2

P = \frac{3}{5}\frac{2}{4}\frac{2}{3} = \frac{2}{10}

N[sub:2oia1avp]1[/sub:2oia1avp] = 1
N[sub:2oia1avp]2[/sub:2oia1avp] = 3

P = \frac{3}{5}\frac{2}{4}\frac{1}{3}\frac{2}{2} = \frac{1}{10}

N[sub:2oia1avp]1[/sub:2oia1avp] = 2
N[sub:2oia1avp]2[/sub:2oia1avp] = 1

P = \frac{2}{5}\frac{3}{4}\frac{2}{3} = \frac{2}{10}

N[sub:2oia1avp]1[/sub:2oia1avp] = 2
N[sub:2oia1avp]2[/sub:2oia1avp] = 2

P = \frac{2}{5}\frac{3}{4}\frac{1}{3}\frac{2}{2} = \frac{1}{10}

N[sub:2oia1avp]1[/sub:2oia1avp] = 3
N[sub:2oia1avp]2[/sub:2oia1avp] = 1

P = \frac{2}{5}\frac{1}{4}\frac{3}{3}\frac{2}{2} = \frac{1}{10}