Answer :

Let E be the event of drawing 2 random balls (a red ball and another black ball.)

E_{1} represents the event of the appearance of 1 or 2 on the die

And E_{2} represents the event of appearance 3,4,5 or 6.

As we need to find the probability of one ball being red and other being black.

But the selection of these balls depends on the number that appeared on the die.

Therefore this problem is of conditional probability, and we need to find P(E).

Now as 2 balls can be selected out of first bag or second which depends on the number appeared on die. So this can be given as-

P(E) = P(E_{1})P(E/E_{1}) + P(E_{2})P(E/E_{2})

As P(E_{1}) = 2/6 = 1/3 {as for die n(s) = 6 and n(E_{1}) = 2}

And P(E_{2}) = 4/6 = 2/3 {as for die n(s) = 6 and n(E_{2}) = 4}

P(E|E_{1}) represents selection of one red and one black ball given that 1 or 2 appeared on die. As per the question now we have to select balls from bag A

∵ bag A contains 10 balls , 2 balls can be selected in ^{10}C_{2} ways.

And 1 black and 1 red ball can be selected in ^{4}C_{1}×^{6}C_{1} ways

∴ P(E|E_{1}) =

Similarly we determine P(E|E_{2})

P(E|E_{2}) =

∴ P(E) =

Hence,

P(E) =

**OR**

Let X be the random variable denoting a number of heads appeared on 4 tosses. So it can take value 0, 1, 2, 3 & 4

Let E denotes the event – head appeared on tossing of a coin

∴ P(E) = 1/2 and P(E’) = 1/2

Now,

Using Binomial distribution and Bernoulli trials:

P(X=0) denotes event of getting 0 head = ^{4}C_{0}(1/2)^{4}

P(X=1) denotes event of getting 1 head = ^{4}C_{1} ×(1/2)^{1}×(1/2)^{3}

P(X=2) denotes event of getting 2 heads = ^{4}C_{2} ×(1/2)^{2}×(1/2)^{2}

P(X=3) denotes event of getting 3 heads = ^{4}C_{3}×(1/2)^{3}×(1/2)^{1}

P(X=4) denotes event of getting 4 heads = ^{4}C_{4}×(1/2)^{4}×(1/2)^{0}

This can be presented in tabular form as shown:

We know that:

Mean = ∑p_{i}x_{i} =

∴ Mean =

And Variance = ∑p_{i}x_{i}^{2} – (Mean)^{2}

σ^{2} = - 2^{2}

∴ Variance =

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