Question

The probability that a person aged 60 years will live up to 70 is 0.65. What is the probability that out of 10 persons aged 60, at least 7 of them will live up to 70?

• A. 0.3158
• B. 0.5318
• C. 0.5138
• D. 0.5831

Hint:

Identify binomial probability problems by the fixed number of independent trials, two possible outcomes (success/failure), and a constant probability of success. The binomial probability formula is key to solving these problems.

Answer 1

The Correct Option is C

This is a binomial probability problem. Let \( p \) be the probability that a person aged 60 lives up to 70, so \( p = 0.65 \). Let \( q \) be the probability that a person aged 60 does not live up to 70, so \( q = 1 - p = 1 - 0.65 = 0.35 \).
We have \( n = 10 \) persons. We want to find the probability that at least 7 of them live up to 70, which means we need to calculate \( P(X \ge 7) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) \), where \( X \) is the number of persons who live up to 70.
The binomial probability formula is \( P(X = k) = \binom{n}{k} p^k q^{n-k} \).
For \( k = 7 \):
$$P(X = 7) = \binom{10}{7} (0.65)^7 (0.35)^3 = 120 \times 0.049175430625 \times 0.042875 \approx 0.25196$$ For \( k = 8 \):
$$P(X = 8) = \binom{10}{8} (0.65)^8 (0.35)^2 = 45 \times 0.03196402990625 \times 0.1225 \approx 0.17628$$
For \( k = 9 \):
$$P(X = 9) = \binom{10}{9} (0.65)^9 (0.35)^1 = 10 \times 0.0207766194390625 \times 0.35 \approx 0.07272$$ For \( k = 10 \):
$$P(X = 10) = \binom{10}{10} (0.65)^{10} (0.35)^0 = 1 \times 0.0135017826353925 \times 1 \approx 0.01350$$ Now, sum these probabilities:
$$P(X \ge 7) = 0.25196 + 0.17628 + 0.07272 + 0.01350 \approx 0.51446$$ This value is very close to option (3) 0.5138. The slight difference might be due to rounding in intermediate steps.