Question

Bag A has 4 green and 6 red shirts while bag B has 2 green and 3 red shirts. What is the probability that the shirt is taken from bag B, knowing that the shirt is green?
1. \(1\over 4\)
2. \(1\over 3\)
3. \(1\over 2\)
4. \(2\over 3\)
5. None of these

Answer 1

Correct Answer - Option 3 : \(1\over 2\)

Concept: 

General Rule:

• The number of ways for selecting r from a group of n (n > r) = nCr 
• The probability of particular case = \(\rm \text{Number of ways for the case can be executed}\over{\text{Total number of ways for selection}}\)

For any two independent events A and B, if P(A) and P(B) is their probability of occurring then:

• P(A∩B) = P(A) × P(B)
• P(A∪B) = P(A) + P(B) - P(A∩B)
• P(A') = 1 - P(A)
• P(B') = 1 - P(B)

Conditional Probability:

If there are two cases A and B having a probability of P(A) and P(B), then,

Probability of A happens given that B definitely has happened P(A|B) = \(\rm P(A\;∩\; B)\over P( B)\)

 

Calculation:

In the given case 

Case A: Selecting the bag and there are 2 bags having an equal possibility of getting choose 

∴ P(A) = \(1\over2 \)

Case B: Selection of the green shirt from any bag

∴ P(B) = \({4\over 10} + {2\over 5}\) = \(4\over 5\)

Now P(A ∩ B) = P(A) × P(B)

⇒ P(A ∩ B) = \(1\over2 \) × \(4\over 5\) = \(2\over 5\)

Probability of case A happens given that case B definitely has happened

P(A|B) = \(\rm P(A\;∩ \;B)\over P( B)\)

⇒ P(A|B) = \({2\over5}\over{4\over5}\) = \(\boldsymbol{1\over2 }\)