In a box of thirty AAA batteries, there are three defective batteries. Two batteries are randomly selected and tested. What is the probability that both are defective, if the first one is not replaced after being tested?
3/30 * 2/29 = ?
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To solve this problem, we can use the concept of conditional probability.
Given that there are 30 AAA batteries and 3 of them are defective, the probability of selecting a defective battery on the first draw is:
P(Defective on first draw) = 3/30 = 1/10
Since the first battery is not replaced, there are now 29 batteries remaining, with 2 defective batteries. So, the probability of selecting a defective battery on the second draw, given that the first battery was defective, is:
P(Defective on second draw | Defective on first draw) = 2/29
To find the probability that both batteries are defective, we need to multiply the probabilities of the individual events together:
P(Both batteries defective) = P(Defective on first draw) * P(Defective on second draw | Defective on first draw)
P(Both batteries defective) = (1/10) * (2/29)
P(Both batteries defective) = 2/290
Therefore, the probability that both batteries are defective, if the first one is not replaced after being tested, is 2/290.
To find the probability of both batteries being defective, given that the first one is not replaced after being tested, we need to calculate the probability of two independent events happening consecutively.
Let's break down the problem step by step:
1. Determine the total number of ways two batteries can be selected from the box of thirty AAA batteries. This can be calculated using combinations. The formula for combinations is:
nCr = n! / (r!(n-r)!),
where n represents the total number of items (in this case, 30 batteries) and r represents the number of items being selected (in this case, 2 batteries).
Therefore, the total number of ways to select two batteries is:
30C2 = 30! / (2!(30-2)!) = 30! / (2!28!) = (30 * 29) / (2 * 1) = 435.
2. Determine the number of ways to select two defective batteries from the three defective batteries. Since there are three defective batteries in the box, the number of ways to select two defective batteries is simply:
3C2 = 3! / (2!(3-2)!) = 3! / (2!1!) = (3 * 2) / (2 * 1) = 3.
3. Finally, calculate the probability by dividing the number of favorable outcomes (two defective batteries) by the total number of possible outcomes (two batteries selected):
Probability = Number of favorable outcomes / Total number of possible outcomes
Probability = (number of ways to select two defective batteries) / (number of ways to select two batteries)
Probability = 3 / 435.
After simplifying, we find that the probability that both batteries are defective, given that the first one is not replaced, is approximately 0.0069 (rounded to four decimal places).