1.) In a shipment of alarm clocks, the probability that one alarm clock is defective is 0.04. Charlie selects three alarm clocks at random. If he puts each clock back with the rest of the shipment before selecting the next one, what is the probability that all three alarm clocks are defective?
2.) A builder has 8 lots available for sale.
·Six lots are greater than one acre.
·Two lots are less than one acre.
What is the probability that the next three lots sold will be greater than one acre?
3.) A cafeteria has 5 turkey sandwiches, 6 cheese sandwiches, and 4 tuna sandwiches. There are two students in line and each will take a sandwich. What is the probability that the first student takes a cheese sandwich and the next student takes a turkey sandwich? 2/15
Wait actually nevermind haha I figured them out:)
Nice, good job.
1.) To solve this problem, we can use the concept of independent events. The probability that one alarm clock is defective is given as 0.04. Since Charlie selects the alarm clocks one at a time and puts each clock back before selecting the next one, we can consider each selection as an independent event.
The probability of selecting a defective alarm clock on the first try is 0.04. The probability of selecting a defective alarm clock on the second try is also 0.04, and the same for the third try. Since each selection is independent, we can multiply these probabilities together to find the probability of all three alarm clocks being defective.
P(all three alarm clocks are defective) = P(defective on first try) * P(defective on second try) * P(defective on third try)
= 0.04 * 0.04 * 0.04
= 0.0016
Therefore, the probability that all three alarm clocks are defective is 0.0016.
2.) In this problem, we need to find the probability that the next three lots sold will be greater than one acre. Out of the 8 lots available for sale, 6 are greater than one acre, and 2 are less than one acre. Since each sale is independent, we can calculate the probability of each sale individually and then multiply them together.
The probability of selling a lot greater than one acre on the first sale is 6/8. Similarly, the probability of selling a lot greater than one acre on the second and third sales is also 6/8 each.
P(next three lots sold will be greater than one acre) = P(sold greater than one acre on first sale) * P(sold greater than one acre on second sale) * P(sold greater than one acre on third sale)
= (6/8) * (6/8) * (6/8)
= 0.421875
Therefore, the probability that the next three lots sold will be greater than one acre is approximately 0.421875.
3.) To find the probability that the first student takes a cheese sandwich and the next student takes a turkey sandwich, we need to consider the total number of possible sandwich combinations and the number of combinations that meet our desired condition.
The total number of possible sandwich combinations is the product of the number of options for the first student (6 cheese sandwiches) and the number of options for the second student (5 turkey sandwiches).
The number of combinations where the first student takes a cheese sandwich and the second student takes a turkey sandwich is the product of the number of options for the first student (6 cheese sandwiches) and the number of options for the second student (5 turkey sandwiches).
Therefore, the probability that the first student takes a cheese sandwich and the next student takes a turkey sandwich is:
P(first student takes a cheese sandwich and next student takes a turkey sandwich) = (6/15) * (5/14)
= 2/15
Therefore, the probability that the first student takes a cheese sandwich and the next student takes a turkey sandwich is 2/15.