500bottles contains 10 bottles that are defective.3bottles are selected,random,without replacement.a)what is the probability that the third one selected is defective given that the first was okay and second one selected was defective?b)what is the probability that the first one selected was defective and the second one selected was okay? c)what is the ptobability that exactly 1 defective bottle is selected?
there are 490 perfect bottles
a- (490/500)*(10/499)*(9/498)
b-(10/500)*(490/499)*(9/498)
c-(10/500*490/499*489/498)+(490/500*10/499*489/498)+(490/500*10/499*489/498)
To solve these probability questions, we need to use the concept of conditional probability and combinations. Let's tackle each question step by step:
a) The probability that the third bottle is defective, given that the first was okay and the second one selected was defective, can be calculated using conditional probability. We start by determining the number of "successful" outcomes and the total number of possible outcomes.
Successful outcomes: In this case, the first bottle is okay, the second bottle is defective, and the third bottle is defective. Since there are 10 defective bottles and 490 non-defective bottles, the number of successful outcomes is 490 * 10 * 9.
Total possible outcomes: We select 3 bottles without replacement, so the total number of possible outcomes is given by the combination formula: C(500, 3) = 500! / (3!(500-3)!).
Therefore, the probability can be calculated as:
P(third bottle defective | first okay and second defective) = [490 * 10 * 9] / [500! / (3!(500-3)!)].
b) The probability that the first bottle is defective and the second one selected is okay can be calculated in a similar way. For this case:
Successful outcomes: The first bottle is defective, the second bottle is non-defective, and any third bottle can be selected. The number of successful outcomes is 10 * (490 * 10) * C(490, 1).
Total possible outcomes: We again select 3 bottles without replacement, so the total number of possible outcomes is C(500, 3).
Therefore, the probability can be calculated as:
P(first bottle defective and second okay) = [10 * (490 * 10) * C(490, 1)] / [C(500, 3)].
c) The probability that exactly 1 defective bottle is selected can be determined by considering the two possible cases: (1) the first bottle is defective and the next two are okay, or (2) the first bottle is okay, the second bottle is defective, and the third one is okay.
For the first case:
Successful outcomes: The first bottle is defective, the second and third are non-defective. The number of successful outcomes is 10 * (490 * 489).
For the second case:
Successful outcomes: The first and third bottles are okay, and the second one is defective. The number of successful outcomes is (490 * 10) * (490).
Total possible outcomes: C(500, 3).
Therefore, the probability can be calculated as:
P(exactly 1 defective) = [(10 * (490 * 489)) + ((490 * 10) * (490))] / [C(500, 3)].
By substituting the values and performing the calculations, you can find the required probabilities for each case.