Assume that the probability for rain on a day is 0.4 . An umbrella salesman can earn 400 per day in case of rain on that day will lose 100 per day if there is no rain . The expected earnings (in) per day of the salesman is

Option a: 400. Solution: Probability of Two Successive Draws Without Replacement Total balls = 13 (8 red, 5 white) Step 1: Probability of drawing 3 white balls in the first draw - Number of ways to choose 3 white balls from 5: (5 choose 3) = 10 - Total ways to choose 3 balls from 13: (13 choose 3) = 13 × 12 × 11/3 × 2 × 1 = 286 P(1st draw is White) = 10/286 = 5/143 Step 2: Probability of drawing 3 red balls in the second draw After the first draw, 10 balls remain (8 red, 2 white). - Number of ways to choose 3 red balls from 8: (8 choose 3) = 8 × 7 × 6/6 = 56 - Total ways to choose 3 balls from 10: (10 choose 3) = 10 × 9 × 8/6 = 120 P(2nd draw is Red | 1st draw is White) = 56/120 = 7/15 Step 3: Joint Probability P = 10/286 × 56/120 = 70/4290 = 7/429 Hence, Option A is the correct answer.

ProbabilityMCQMTP June 22Question 3411 of 187

All Questions

Assume that the probability for rain on a day is $\displaystyle 0.4$ . An umbrella salesman can earn $\displaystyle 400$ per day in case of rain on that day will lose $\displaystyle 100$ per day if there is no rain . The expected earnings (in) per day of the salesman is

Options

\displaystyle 400

\displaystyle 200

\displaystyle 100

\displaystyle 0

For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option a — $\displaystyle 400$

All Options:

A $\displaystyle 400$
B $\displaystyle 200$
C $\displaystyle 100$
D $\displaystyle 0$

Detailed Solution & Explanation

**Probability of Two Successive Draws Without Replacement** Total balls = 13 (8 red, 5 white) **Step 1: Probability of drawing 3 white balls in the first draw** - Number of ways to choose 3 white balls from 5:

\binom{5}{3} = 10

- Total ways to choose 3 balls from 13:

\binom{13}{3} = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = 286

P(\text{1st draw is White}) = \frac{10}{286} = \frac{5}{143}

**Step 2: Probability of drawing 3 red balls in the second draw** After the first draw, 10 balls remain (8 red, 2 white). - Number of ways to choose 3 red balls from 8:

\binom{8}{3} = \frac{8 \times 7 \times 6}{6} = 56

- Total ways to choose 3 balls from 10:

\binom{10}{3} = \frac{10 \times 9 \times 8}{6} = 120

P(\text{2nd draw is Red} | \text{1st draw is White}) = \frac{56}{120} = \frac{7}{15}

**Step 3: Joint Probability**

P = \frac{10}{286} \times \frac{56}{120} = \frac{70}{4290} = \frac{7}{429}

Hence, **Option A** is the correct answer.

About This Chapter: Probability

Paper

Paper 3: Quantitative Aptitude

Weightage

5-7 Marks

Key Topics

Probability Operations, Expected Value

A logic-heavy chapter dealing with random experiments, events (mutually exclusive, exhaustive), set theory probability, conditional probability, and Bayes' Theorem. It forms the basis for Theoretical Distributions.

View Official ICAI Syllabus

Exam Strategy Tip

Always draw a quick Venn Diagram or tree when faced with 'At least one' or 'Only A but not B' wording. It saves you from double-counting.

Key Concepts to Understand

Probability

A numerical measure of the likelihood of an event occurring, expressed as a number between 0 (impossible) and 1 (certain). Formula: P(A) = Number of favorable outcomes / Total number of equally likely outcomes.

More Questions from Probability

If $\displaystyle P(A) = \frac{1}{2}$ , $\displaystyle P(B) = \frac{1}{3}$ , $\displaystyle P(A \cap B) = \frac{1}{4}$ , then the value of $\displaystyle P(A' \cup B')$ is

Two dice are thrown simultaneously. Find the probability that the sum of digits on the two dice would be 8 or more.

A number is selected from the first 20 natural numbers. Find the probability that it would be divisible by 3 or 7.

A father had three sons namely, Kailash, Harish and Prakash. All are above 65 years in age. Prakash happens to be the eldest while Kailash as youngest. As per the health history, it is estimated that the probability that Kailash survives another 5 years is $\displaystyle \frac{4}{5}$ , Harish survives another 5 years is $\displaystyle \frac{3}{5}$ and Prakash survives another 5 years is $\displaystyle \frac{1}{2}$ . The probabilities that Kailash and Harish survive another 5 years is 0.46, Harish and Prakash survive another 5 years is 0.32 and Kailash and Prakash survive another 5 years is 0.48. The probability that all three sons survive another 5 years is 0.26. What shall be the probability that at least one of them survives another 5 years?

In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected is

Two cards are drawn from a pack of 52 cards. The probability that one is a spade and one is a heart; is

Ready to Master Probability?

Practice all 187 questions with instant feedback, earn XP, track your streaks, and ace your CA Foundation exam.

Start Practicing — It's Free

Assume that the probability for rain on a day is 0.4\displaystyle 0.40.4. An umbrella salesman can earn 400\displaystyle 400400 per day in case of rain on that day will lose 100\displaystyle 100100 per day if there is no rain . The expected earnings (in) per day of the salesman is