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) = \({2\over5}\over{4\over5}\) = \(\boldsymbol{1\over2 }\)