a city law-enforcement official has stated that 20% of the items sold by pawn shops within the city have been stole. Ralph has just purchased 4 items from one of the city's pawn shops. Assuming that the official is correct, and for x = the number of ralph's purchases that have been stolen, determine the following:
A) P(x)
B) P (2< or equal to x)
C) P( 1 < or equal to x < or equal to 3)
D) P (x < or equal to 2)
To determine the probabilities, we can use the concept of probability distributions. In this case, we will be using the binomial probability distribution, as we have a fixed number of trials (Ralph's purchases) with two possible outcomes (stolen or not stolen) and a known probability of success (20%).
A) P(x): The probability of exactly x items out of four being stolen.
To calculate this, we need to use the formula for binomial probability:
P(x) = C(n, x) * p^x * (1 - p)^(n - x)
Where:
- C(n, x) represents the combination value of selecting x items out of n.
- p is the probability of success (in this case, p = 0.2 since 20% of items are stolen).
- n is the total number of trials (in this case, n = 4).
- x is the number of successful trials (the number of stolen items by Ralph).
Let's calculate the probabilities for different x values:
P(0) = C(4, 0) * (0.2)^0 * (0.8)^(4 - 0) = 1 * 1 * 0.8^4 = 0.4096
P(1) = C(4, 1) * (0.2)^1 * (0.8)^(4 - 1) = 4 * 0.2 * 0.8^3 = 0.4096
P(2) = C(4, 2) * (0.2)^2 * (0.8)^(4 - 2) = 6 * 0.04 * 0.8^2 = 0.1536
P(3) = C(4, 3) * (0.2)^3 * (0.8)^(4 - 3) = 4 * 0.008 * 0.8^1 = 0.0512
P(4) = C(4, 4) * (0.2)^4 * (0.8)^(4 - 4) = 1 * 0.0016 * 0.8^0 = 0.0016
Therefore, P(x) for each value of x is as follows:
P(0) = 0.4096, P(1) = 0.4096, P(2) = 0.1536, P(3) = 0.0512, P(4) = 0.0016.
B) P(2 ≤ x): The probability of getting at least 2 stolen items.
To calculate this, we need to sum the probabilities for x = 2, 3, and 4.
P(2 ≤ x) = P(2) + P(3) + P(4) = 0.1536 + 0.0512 + 0.0016 = 0.2064
Therefore, P(2 ≤ x) is 0.2064.
C) P(1 ≤ x ≤ 3): The probability of getting 1, 2, or 3 stolen items.
To calculate this, we need to sum the probabilities for x = 1, 2, and 3.
P(1 ≤ x ≤ 3) = P(1) + P(2) + P(3) = 0.4096 + 0.1536 + 0.0512 = 0.6144
Therefore, P(1 ≤ x ≤ 3) is 0.6144.
D) P(x ≤ 2): The probability of getting 0, 1, or 2 stolen items.
To calculate this, we need to sum the probabilities for x = 0, 1, and 2.
P(x ≤ 2) = P(0) + P(1) + P(2) = 0.4096 + 0.4096 + 0.1536 = 0.9728
Therefore, P(x ≤ 2) is 0.9728.