7/17/2019 Answers of Tutroial Sheet 4

http://slidepdf.com/reader/full/answers-of-tutroial-sheet-4 1/1

Answer to Tutorial Sheet-4

1. (a) M X1+X2(t) =  e−λ1(e

t−1)e−λ2(e

t−1) since X 1  and  X 2  are independent.

⇒ M X1+X2(t) =  e−(λ1+λ2)(e

t−1).  Hence follows poisson distribution. Similar proof for (ii).

(b) Both statements are not true.

2.   P (X 1 + X 2 >  5) = 1 − P (X 1 + X 2 ≤ 5) = 1 −5

x=0

e−88x

x!   = 0.8088.

3. (a) P (X  = 2) =   e−1.2143(1.21432

2!  = 0.2189.

(b) P (X  ≥ 1) = 1 − P (= 0) = 1 − e−0.48572 = 0.3847.

4. First show that P (X > n) =  q n. Now  P (X > m + n|X > m) =   P (X>m+n)P (X>m)

  = q m+n/q m = q n. Hence

proved.

5. Expected profit=E (T ) = 600t − E (50X 2) = 600t − 50(0.8t + 0.8t2) and it attains maximum att = 8.75.

6. First show that   P (X   =   r|(X  + y) =   n) =   nC r prq n−r,  where   p   =   λ1

λ1+λ2and   q   =   λ2

λ1+λ2.  Hence

proved.

7.   P (X  = r|X  + Y   = k) =   1k−1

; r = 1, 2, · · ·  , (k − 1).  Hence follows uniform distribution.

8. Clearly the random variable follows geometric distribution. So you need to maximize the densityfunction with respect to  r   for  x  = 5.  The maximum value attains at r  = 0.8.

9.   P (X  ≤ 2) =2

x=0e−λ

λx

x!  ,  where λ  =   25

10.

*********** THE END ************

1