Question

A bag contains 4 balls. Two balls are drawn at random and are found to be white. What is the probability that minimum balls are white?

Answer 1

as 2 balls are white
therefore ,
4-2=2
p(of getting non white ball)=2/4
=1/2

Answer 2

Answer:

= $\frac{3}{5}$

Step-by-step explanation:

Let us define the following events

       E: drawn balls are white      ,            A: 2 white balls in bag

       B: 3 white balls in bag         ,           C: 4 white balls in bag

The, P(A) = P(B) = P(C) = $\frac{1}{3}$

and     P$(\frac{E}{A} )$ = ²C₂ / ⁴C₂ = 1/6

       P(E/B) = ³C₂ / ⁴C₂ = 3/6

       P(E/C) = ⁴C₂/⁴C₂ = 1

Now, By applying Baye's Theorem

       P(C/A) = P(C) . P(E/C) / P(A) . P(E/A) + P(B) . P(E/B) + P(C) P(E/C)

 = 1/3 × 1 / ( 1/3 × 1/6 ) + (1/3 × 3/6) + (1/3 × 1)

 = 1 / 1/6 + 3/6 + 1

 = $\frac{3}{5}$

GOOD LUCK !!