Question:
A bag contains \(4\) red and \(3\) black balls. A second bag contains \(2\) red and \(3\) black balls. One bag is selected at random. If from the selected bag, one ball is drawn at random, then the probability that the ball drawn is red is
A bag contains \(4\) red and \(3\) black balls. A second bag contains \(2\) red and \(3\) black balls. One bag is selected at random. If from the selected bag, one ball is drawn at random, then the probability that the ball drawn is red is
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When an experiment occurs in stages, use the law of total probability:
\[
P(A)=P(B_1)P(A\mid B_1)+P(B_2)P(A\mid B_2).
\]
Multiply along branches and then add the results.
Updated On: Jun 22, 2026
- \(\frac{39}{70}\)
- \(\frac{41}{70}\)
- \(\frac{29}{70}\)
- \(\frac{17}{35}\)
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The Correct Option is D
Solution and Explanation
Step 1: Define the events.
Let: \[ B_1=\text{selection of first bag} \] \[ B_2=\text{selection of second bag} \] Since one bag is selected at random, \[ P(B_1)=P(B_2)=\frac12 \] Let \[ R=\text{event of drawing a red ball} \]
Step 2: Find probability of drawing a red ball from each bag.
For the first bag: \[ P(R\mid B_1)=\frac47 \] For the second bag: \[ P(R\mid B_2)=\frac25 \]
Step 3: Apply the law of total probability.
\[ P(R) = P(B_1)P(R\mid B_1) + P(B_2)P(R\mid B_2) \] Substituting values, \[ P(R) = \frac12\cdot\frac47 + \frac12\cdot\frac25 \] \[ = \frac27+\frac15 \]
Step 4: Simplify the expression.
Taking LCM \(35\), \[ P(R) = \frac{10}{35}+\frac{7}{35} \] \[ = \frac{17}{35} \]
Step 5: Final conclusion.
Therefore, the probability that the ball drawn is red is \[ \boxed{\frac{17}{35}} \]
Let: \[ B_1=\text{selection of first bag} \] \[ B_2=\text{selection of second bag} \] Since one bag is selected at random, \[ P(B_1)=P(B_2)=\frac12 \] Let \[ R=\text{event of drawing a red ball} \]
Step 2: Find probability of drawing a red ball from each bag.
For the first bag: \[ P(R\mid B_1)=\frac47 \] For the second bag: \[ P(R\mid B_2)=\frac25 \]
Step 3: Apply the law of total probability.
\[ P(R) = P(B_1)P(R\mid B_1) + P(B_2)P(R\mid B_2) \] Substituting values, \[ P(R) = \frac12\cdot\frac47 + \frac12\cdot\frac25 \] \[ = \frac27+\frac15 \]
Step 4: Simplify the expression.
Taking LCM \(35\), \[ P(R) = \frac{10}{35}+\frac{7}{35} \] \[ = \frac{17}{35} \]
Step 5: Final conclusion.
Therefore, the probability that the ball drawn is red is \[ \boxed{\frac{17}{35}} \]
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