Question

If X and Y are two events such that probability of X is 1/2 , probability of Y is k, probability of occurrence of at least of one of the two events X and Y is 4/5 . For what value of k (i) X and Y are disjoint (ii) X and Y are independent

Answer 1

$\huge\bold\orange{Solution}$

Let's x and y be two independent integers -value random variables with distribution

functions

$m {1} (x) \: and \: m2(x)$

Respectively.

Then the convolution of m1(x) and m2(x) is the distribution function m3=m1∗m2 given by

$function m3=m1∗m2m3=m1∗m2 given by</p><p></p><p>m3(j)=∑km1(k)⋅m2(j−k),(7.1.3)(7.1.3)m3(j)=∑km1(k)⋅m2(j−k),</p><p></p><p>for j = . . . , −2, −1, 0, 1, 2, . . .. The function m3(x) is the distribution function of the random variable Z = X + Y.$

for j = . . . , −2, −1, 0, 1, 2, . . .. The function m3(x) is the distribution function of the random variable Z = X + Y.

It is easy to see that the convolution operation is commutative, and it is straightforward to show that it is also associative.

Now let Sn=X1+X2+...+Xn be the sum of n independent random variables of an independent trials process with common distribution function m defined on the integers. Then the distribution function of S1 is m. We can write

$Sn=Sn−1+Xn(7.1.4)(7.1.4)Sn=Sn−1+Xn</p><p></p><p>Thus, since we know the distribution function of XnXn is m, we can find the distribution function of SnSn by induction.$

✍Hope it's helpful to you ✍

Answer 2

$\huge{\frak{Question}}$

If X and Y are two events such that probability of X is 1/2 , probability of Y is k, probability of occurrence of at least of one of the two events X and Y is 4/5 . For what value of k

(i) X and Y are disjoint

(ii) X and Y are independent

$\huge{\frak{Answer}}$

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