A bag contains 4 red and 6 blue balls. Two balls are drawn one... | Mathematics MCQ | Examoogle

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A bag contains 4 red and 6 blue balls. Two balls are drawn one by one without replacement. What is the probability that both balls are red?

4/25

2/15

1/5

3/10

Correct Answer

⚙️ TE • Technical Conditional ProbabilityMathematicsProbability

Option B

2/15

Quick Summary:

Use the product of ratios: $\frac{4}{10} \times \frac{3}{9} = \frac{2}{5} \times \frac{1}{3} = \frac{2}{15}$ . Mentally reduce fractions before multiplying.

📐MAMath SolutionConditional Probability

📋 Given

Total red balls = 4, Total blue balls = 6, Total balls = 10. Two balls drawn one by one without replacement.

🔢 Formula Used

P(R1∩R2)=P(R1)×P(R2∣R1)=n(R)n(S)×n(R)−1n(S)−1P(R_1 \cap R_2) = P(R_1) \times P(R_2 | R_1) = \frac{n(R)}{n(S)} \times \frac{n(R)-1}{n(S)-1}P(R1​∩R2​)=P(R1​)×P(R2​∣R1​)=n(S)n(R)​×n(S)−1n(R)−1​

⚡ Exam Hall Shortcut / Speed Trick

Use the product of ratios: $\frac{4}{10} \times \frac{3}{9} = \frac{2}{5} \times \frac{1}{3} = \frac{2}{15}$ . Mentally reduce fractions before multiplying.

⚠️ Common Student Trap / Pitfall

Students often forget to decrease both the numerator (red balls) and the denominator (total balls) for the second draw since it is without replacement.

📊 Diagram / Illustration

🔢 Step-by-Step Solution

Calculate probability of first red ball

The probability of drawing the first red ball out of the total 10 balls is given by the ratio of red balls to the total number of balls.

P(R1)=410=25P(R_1) = \frac{4}{10} = \frac{2}{5}P(R1​)=104​=52​

Calculate probability of second red ball

Since one red ball is already removed, there are now 3 red balls left and 9 total balls remaining in the bag.

P(R2∣R1)=39=13P(R_2 | R_1) = \frac{3}{9} = \frac{1}{3}P(R2​∣R1​)=93​=31​

Calculate combined probability

Apply the multiplication rule for dependent events by multiplying the probability of the first event by the conditional probability of the second.

P(R1∩R2)=25×13=215P(R_1 \cap R_2) = \frac{2}{5} \times \frac{1}{3} = \frac{2}{15}P(R1​∩R2​)=52​×31​=152​

✅

B is correct because the product of the sequential probabilities $\frac{4}{10} \times \frac{3}{9}$ results in $\frac{2}{15}$ .

Core Concepts Used

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Dependent Events Conditional Probability Multiplication Rule

💡 EXAM TIP

This logic is essential for Permutation and Combination problems involving selection without replacement, frequently seen in banking exams.

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