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# An electronics store has received a shipment of 20 table radios that have connections for an iPod or iPhone. Twelve of these have two slots

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An electronics store has received a shipment of 20 table radios that have connections for an iPod or iPhone. Twelve of these have two slots (so they can accommodate both devices), and the other eight have a single slot. Suppose that eight of the 20 radios are randomly selected to be stored under a shelf where the radios are displayed, and the remaining ones are placed in a storeroom. Let X = the number among the radios stored under the display shelf that have two slots.Compute P(X = 2), P(X ≤ 2), and P(X ≥ 2)

Guest Mar 2, 2015

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The probability that X = 2 is given by

C(12,2)* C(8,6) / C(20,8) = about 1.47%

The probability that X ≤ 2 = probability that none, one or two of the two slot models are chosen =

C(12,0)* C(8,8) / C(20,8) + C(12,1)* C(8,7) / C(20,8) + C(12,2)* C(8,6) / C(20,8) = about 1.54%

The probabliity that X ≥ 2 = 1 minus the probability that none or one of the two slot models are chosen =

1 - [C(12,0)* C(8,8) / C(20,8) + C(12,1)* C(8,7) / C(20,8)]  = about 99%

CPhill  Mar 2, 2015
Sort:

#1
+85726
+5

The probability that X = 2 is given by

C(12,2)* C(8,6) / C(20,8) = about 1.47%

The probability that X ≤ 2 = probability that none, one or two of the two slot models are chosen =

C(12,0)* C(8,8) / C(20,8) + C(12,1)* C(8,7) / C(20,8) + C(12,2)* C(8,6) / C(20,8) = about 1.54%

The probabliity that X ≥ 2 = 1 minus the probability that none or one of the two slot models are chosen =

1 - [C(12,0)* C(8,8) / C(20,8) + C(12,1)* C(8,7) / C(20,8)]  = about 99%

CPhill  Mar 2, 2015

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