Question

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)

Answer 1

This problem involves using the hypergeometric probability distribution to calculate P(X = 2), P(X ≤ 2), and P(X ≥ 2) for radios with two slots being selected to place under a store display. The calculations require combinations and understanding cumulative probabilities.

Step-by-step explanation:

This is a question of hypergeometric probability distribution since we're dealing with selections without replacement from two different groups (radios with two slots and radios with one slot).

Calculating P(X = 2)

The probability that X, the number of radios with two slots on the display shelf, equals 2 can be found with the formula:

P(X = k) = [(C(Group1, k) * C(Group2, n - k)) / C(Total, n)]

Where:

P(X = 2) = [(C(12, 2) * C(8, 6)) / C(20, 8)]

Calculating P(X ≤ 2)

P(X ≤ 2) is the cumulative probability that X is less than or equal to 2. This encompasses P(X = 0), P(X = 1), and P(X = 2).

P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)

Calculating P(X ≥ 2)

P(X ≥ 2) is the cumulative probability that X is greater than or equal to 2. This encompasses P(X = 2), P(X = 3), ..., P(X = 8).

Since the probabilities must sum to 1:

P(X ≥ 2) = 1 - [P(X = 0) + P(X = 1)]