A bag contains 4 red balls and 5 blue balls. Two balls are drawn at r

1.	A bag contains 4 red balls and 5 blue balls. Two balls are drawn at random without replacement. If the first ball drawn is blue, what is the probability the second ball is also blue?1. \(\frac 5 9\)2. \(\frac 1 2\)3. \(\frac {5}{18}\)4. \(\frac 4 9\)
Answer» Correct Answer - Option 2 : \(\frac 1 2\) Concept: If S is a sample space and A is a favourable event then the probability of A is given by: \(\rm P(E)=\dfrac{n(A)}{n(S)}\) Calculation: Total balls in bag = 4 + 5 = 9 balls Given: Two balls are drawn at random without replacement After the first ball is drawn and found to be blue, there are now 8 balls left in the bag, 4 of which are blue. Probability the second ball is blue = \(\frac 4 8 = \frac 1 2\)

A bag contains 4 red balls and 5 blue balls. Two balls are drawn at random without replacement. If the first ball drawn is blue, what is the probability the second ball is also blue?1. \(\frac 5 9\)2. \(\frac 1 2\)3. \(\frac {5}{18}\)4. \(\frac 4 9\)

Answer» Correct Answer - Option 2 : \(\frac 1 2\)

Concept:

If S is a sample space and A is a favourable event then the probability of A is given by:

\(\rm P(E)=\dfrac{n(A)}{n(S)}\)

Calculation:

Total balls in bag = 4 + 5 = 9 balls

Given: Two balls are drawn at random without replacement

After the first ball is drawn and found to be blue, there are now 8 balls left in the bag, 4 of which are blue.

Probability the second ball is blue = \(\frac 4 8 = \frac 1 2\)

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A bag contains 4 red balls and 5 blue balls. Two balls are drawn at random without replacement. If the first ball drawn is blue, what is the probability the second ball is also blue?1. \(\frac 5 9\)2. \(\frac 1 2\)3. \(\frac {5}{18}\)4. \(\frac 4 9\)

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