A sample of 4 different calculators is randomly selected from a group containing 17 that are defective and 36 that have no defects. What is the probability that at least one of the calculators is defective?

17/36

Find P(0), then take 1 - P(0) for your probability (since the problem says "what is the probability that at least one" is defective). Using the binomial probability function (you can use a binomial probability table as well):

P(x) = (nCx)(p^x)[q^(n-x)]
n = 4
x = 0
p = 17/53 = .32 (note: 53 = 17 + 36)
q = 1 - p = .68
Substituting:
P(0) = (4C0)(.32^0)(.68^4) = .2138
Now, take 1 - .2138 -->this will be your probability.

I hope this helps.

To find the probability that at least one calculator is defective, we can calculate the complement of the event that none of the calculators are defective.

The total number of calculators in the group is 17 defective + 36 non-defective = 53 calculators.

To find the probability that none of the calculators are defective, we need to select all 4 calculators from the 36 that have no defects.

The probability that the first calculator selected is non-defective is 36/53.
The probability that the second calculator selected is non-defective is 35/52 (since there is one less non-defective calculator remaining).
The probability that the third calculator selected is non-defective is 34/51.
The probability that the fourth calculator selected is non-defective is 33/50.

Now, we calculate the probability of none of the calculators being defective by multiplying all these probabilities together:

(36/53) * (35/52) * (34/51) * (33/50) ≈ 0.4376

Finally, we calculate the probability that at least one calculator is defective by subtracting the probability of none of the calculators being defective from 1:

P(at least one defective) = 1 - 0.4376 ≈ 0.5624

So, the probability that at least one of the calculators is defective is approximately 0.5624 or 56.24%.

To find the probability that at least one of the calculators is defective, we need to find the probability of the complement event (i.e., none of the calculators being defective) and subtract it from 1.

To calculate the probability that none of the calculators is defective, we need to consider the number of ways we can choose 4 calculators from the group of 36 non-defective calculators divided by the total number of ways we can choose 4 calculators from the entire group.

The number of ways to choose 4 calculators from a set of 36 calculators with no defects can be calculated using the combination formula: C(36, 4) = 36! / (4! * (36-4)!).

The total number of ways to choose 4 calculators from the entire group (including both defective and non-defective calculators) can be calculated using the combination formula: C(53, 4) = 53! / (4! * (53-4)!).

Therefore, the probability of none of the calculators being defective is C(36, 4) / C(53, 4).

Finally, subtracting this probability from 1 gives us the probability that at least one of the calculators is defective:

P(at least one defective calculator) = 1 - (C(36, 4) / C(53, 4)).

Now, you can plug the numbers into the formula and calculate the probability.