The probability that a man who is 85 years old will die before attaining the age of 90 is 1/3. A_1,A_2,A_3,A_4 are four persons aged 85 years. The probability that A_1 will die before attaining 90 and will be the first to die is:

C. 65/324. Step 1: If A_1 and exactly j of the other three die before 90, then by symmetry A_1 is the first of these j+1 deaths with probability 1/j+1. Step 2: P(A_1 dies and exactly j others die)=1/3(3/j)(1/3)^j(2/3)^3-j, i.e. 1/3⋅8/27,1/3⋅12/27,1/3⋅6/27,1/3⋅1/27 for j=0,1,2,3. Step 3: P=1/3[8/27+12/27⋅1/2+6/27⋅1/3+1/27⋅1/4]=65/324. Check: At least one of the four dies with probability 1-(2/3)^4=65/81, and each of the four is equally likely to be the first, so P=1/4⋅65/81=65/324. Trap: (B) 13/81 divides 65/81 by 5 instead of 4; there are only four persons.

Question 28 of 63 · Probability

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The probability that a man who is 85 years old will die before attaining the age of 90 is $\frac{1}{3}$ . $A_{1}, A_{2}, A_{3}, A_{4}$ are four persons aged 85 years. The probability that $A_{1}$ will die before attaining 90 and will be the first to die is:

Answer (Detailed Solution Below)View answer ▾

Answer: Option C — $\frac{65}{324}$

Detailed Solution

If $A_{1}$ and exactly $j$ of the other three die before 90, then by symmetry $A_{1}$ is the first of these $j + 1$ deaths with probability $\frac{1}{j + 1}$ .

$P (A_{1} d i es an d e x a c tl y j o t h er s d i e) = \frac{1}{3} (\frac{3}{j}) (\frac{1}{3})^{j} (\frac{2}{3})^{3 - j}$ , i.e. $\frac{1}{3} \cdot \frac{8}{27}, \frac{1}{3} \cdot \frac{12}{27}, \frac{1}{3} \cdot \frac{6}{27}, \frac{1}{3} \cdot \frac{1}{27}$ for $j = 0, 1, 2, 3$ .

$P = \frac{1}{3} [\frac{8}{27} + \frac{12}{27} \cdot \frac{1}{2} + \frac{6}{27} \cdot \frac{1}{3} + \frac{1}{27} \cdot \frac{1}{4}] = \frac{65}{324}$ .

✓ Check

At least one of the four dies with probability

1 - (\frac{2}{3})^{4} = \frac{65}{81}

, and each of the four is equally likely to be the first, so

P = \frac{1}{4} \cdot \frac{65}{81} = \frac{65}{324}

⚠ Common trap

(B)

\frac{13}{81}

divides

\frac{65}{81}

by 5 instead of 4; there are only four persons.

Quick tip

$P = \frac{1}{4} [1 - (\frac{2}{3})^{4}] = \frac{65}{324}$

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Probability — Preparation

This question tests your understanding of Probability, a key topic in the Quantitative Aptitude section. The detailed solution above uses a step-by-step approach to show the reasoning, and where available, a shortcut that helps you solve similar questions in under 90 seconds during the actual exam.

Probability measures how likely an event is, as a number between 0 and 1. It equals the number of favourable outcomes divided by the total number of possible outcomes, when all outcomes are equally likely. This topic covers coins, dice, playing cards, drawing balls from a bag, complementary events, independent events, and conditional probability. The most useful tools are the complement rule (P(at least one) = 1 - P(none)) and the multiplication rule for independent events. For engineering and MCA entrance exams, the topic also includes the addition theorem, conditional probability, Bayes' theorem, and probability distributions. A clear sample space is the key to avoiding mistakes, so always list or count all outcomes carefully. Practise the probability questions below. Each comes with a step-by-step solution that shows how the sample space and favourable outcomes are counted.

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