The Vardon Exploration Company is getting ready to leave for South America to explore for oil. One piece of equipment requires 10 batteries that must operate for more than 2 hours. The batteries being used have a 85 percent chance of lasting for 2 hours or more. The exploration leader plans to take 15 batteries. Assuming that the conditions of the binomial apply, the probability that the supply of batteries will contain enough good ones to operate the equipment is:

Question

BlackTom AI · Accepted Answer

We have a binomial scenario: n = 15 trials (batteries), each with probability p = 0.85 of being 'good' (lasting at least 2 hours). The equipment needs 10 good batteries, so we want the probability that the number of successes X is at least 10: P(X ≥ 10).

Option by option analysis:

Option A: 0.9832
- This value represents the probability of obtaining 10 or more good batteries out of 15 when p = 0.85. The exact computation would be P(X ≥ 10) = sum from k=10 to 15 of C(15, k) (0.85)^k (0.15)^(15−k). Given the high success probability per battery and a relatively large number of batteries, the tail probability for at least 10 successes is commonly in the high 0.9s range. Therefore 0.9832 is a plausible exact tail probability for this setup.

Option B: 0.9964
- This is a very high tail probability. While P(X ≥ 10) is indeed large for p = 0.85 and n = 15, 0.9964 would imply an even smaller chance of getting fewer than 10 good batteries (i.e., X ≤ 9) than typical for these parameters. Since there is still a nonzero probability of having 9 or fewer good batteries when p = 0.85, 0.9964 is likely too large for P(X ≥ 10).

Option C: 0.0132
- This value is extremely small, indicating almost no chance of obtaining at least 10 good batteries. With p = 0.85 and 15 trials, the expected number of good batteries is 12.75, and the probability of falling to 9 or fewer successes would be quite small but not near 0.01 for this parameter set. Thus 0.0132 is inconsistent with the binomial distribution in this context.

Option D: 0.0449
- A probability around 0.045 would suggest a modest chance of having at least 10 good batteries, which contradicts the intuition from a high-p binomial with n = 15 (the mean is 12.75, and the probability of getting fewer than 10 is not large). This value underestimates P(X ≥ 10) for these parameters.

Additional notes on the reasoning pattern:
- The core concept is a binomial tail probability, P(X ≥ 10) with X ~ Binomial(n=15, p=0.85).
- You could verify by computing the complementary probability P(X ≤ 9) = sum_{k=0}^{9} C(15, k) (0.85)^k (0.15)^{15−k} and then subtracting from 1 to obtain P(X ≥ 10). This typically yields a value around 0.98 for these parameters.
- When evaluating options, compare how plausible each tail probability is given a high per-trial success rate and a sample size that yields a mean well above 9; options that are far from this expectation (near 0.01 or near 1.0 with no room for the other tail) are less consistent with the binomial behavior described.

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