3 batteries are chosen at random from 15 batteries of which 5 are defective. Find the probability if:
1) none of 3 are defective
2) exactly one is defective
3) 2 defective and 1 non-defective
4) at least one is non-defective
What should I do? Is it permutation or a combination? how do you solve it? Please help me. :(
To solve these probability questions, we can use combinations. Combination is the selection of objects without regard to the order, while permutation takes into account the order in which the objects are selected.
1) To find the probability that none of the 3 chosen batteries are defective, we need to select 3 non-defective batteries from the 10 non-defective batteries available (since out of the 15 batteries, 5 are defective). We can use the combination formula.
The combination formula is:
nCr = (n!)/((r!)(n-r)!)
Where n is the total number of objects and r is the number of objects being selected.
Using n = 10 (non-defective batteries) and r = 3 (number of batteries being selected), the calculation becomes:
10C3 = (10!)/((3!)(10-3)!)
= (10!)/(3!7!)
= (10 x 9 x 8)/(3 x 2 x 1)
= 120
Since there are 120 ways to select 3 non-defective batteries from a pool of 10, and there are a total of 15 batteries from which 3 are being chosen, the probability that none of the 3 batteries are defective is 120/455.
2) To find the probability that exactly one battery is defective, we need to select 1 defective battery and 2 non-defective batteries.
The combination formula can be used as well. The calculation becomes:
5C1 = (5!)/((1!)(5-1)!)
= (5!)/(1!4!)
= 5
Similarly, selecting 2 non-defective batteries from the remaining 10 non-defective batteries:
10C2 = (10!)/((2!)(10-2)!)
= (10!)/(2!8!)
= 45
Multiplying these calculations together gives us:
P(exactly one defective) = (5C1) * (10C2) = 5 * 45 = 225
Since there are 225 ways to have exactly one defective battery out of the 15 batteries from which 3 are being chosen, the probability is 225/455.
3) To find the probability that 2 batteries are defective and 1 is non-defective, we need to select 2 defective batteries and 1 non-defective battery.
Using the combination formula, the calculation becomes:
5C2 = (5!)/((2!)(5-2)!)
= (5!)/(2!3!)
= 10
Similarly, selecting 1 non-defective battery from the remaining 10 non-defective batteries:
10C1 = (10!)/((1!)(10-1)!)
= (10!)/(1!9!)
= 10
Multiplying these calculations together gives us:
P(2 defective and 1 non-defective) = (5C2) * (10C1) = 10 * 10 = 100
Since there are 100 ways to have 2 defective batteries and 1 non-defective battery out of the 15 batteries from which 3 are being chosen, the probability is 100/455.
4) To find the probability that at least one battery is non-defective, we can find the complement of the event where all 3 batteries are defective and subtract it from 1.
The probability that all 3 batteries are defective is:
P(all 3 defective) = (5C3) * (10C0) = 1 * 1 = 1
Therefore, the probability of at least one non-defective battery is:
P(at least one non-defective) = 1 - P(all 3 defective) = 1 - 1 = 0
Since the probability of selecting at least one non-defective battery is 0, it means that there is no possibility of selecting a set of 3 batteries with at least one non-defective battery from the pool of 15 batteries, given that 5 of them are defective.
Please note that the calculation assumes that each battery is equally likely to be chosen and that the selections are made without replacement (once a battery is chosen, it is not put back into the pool for subsequent selections).