Question

four bad oranges are accidentally mixed with 16 good ones. find the probability distribution of the number of bad oranges when two oranges are drawn at random from this lot.find the mean and variance of the distribution

Answer 1

total number of oranges = 16+4 = 20
Number of Bad oranges = 4

Probability that out of those 2 are bad =   4C2/20
 Thus

Proababiltiy =  4*3/20
                    = 3/5
Thus the probability is 3/5

Answer 2

Hey !!

Answer:

= $\frac{64}{199}$

Step-by-step explanation:

Let E₁ , E₂ , E₃ and A be events such that

E₁ = Bag I is chosen

E₂ = Bag II is chosen

E₃ = Bag III is chosen

A = Two balls drawn from the chosen bag are white and red

P(E₁) = 1/3 , P(E₂) = 1/3 , P(E₃) = 1/3

P(A/E₁) = 1/6 * 3/6 + 3/6 * 1/6

= 3/36 + 3/36

= 6/36

= 1/6

P(A/E₂) = 2/4 * 1/4 + 1/4 * 2/4

            = 2/16 + 2/16

            = 4/16

            = 1/4

P(A/E₃) = 4/9 * 2/9 + 2/9 * 4/9

             = 16/81

We require,

P (E₃ / A) = P(E₃).P(A / E₃) / P(E₁) . P(A / E₁) + P(E₂).P( A / E₂) + P(E₃) . P(A / E₃)

= 1/3 * 16/81 / 1/3 * 1/6 + 1/3 * 1/4 + 1/3 * 16/81

= 16/81 / 1/6 + 1/4 + 16/81

= 16/81 / 54 + 81 + 64 / 27 * 12

= 16/81 * 27 * 12 / 199

= 64 / 199

Good luck !!