2 more questions... ^^'' I found these two kind of difficult... please teach me to solve them...

Attachment 19327

Answer:

Attachment 19328

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- Oct 14th 2010, 01:36 AMcloud5Discrete Probability Distribution III
2 more questions... ^^'' I found these two kind of difficult... please teach me to solve them...

Attachment 19327

Answer:

Attachment 19328 - Oct 14th 2010, 06:14 AMmathaddict
(14) X~B(40 , 1/20) implies X~P(2)

Calculate P(X>=3)

The rest of the two parts are cases of binomial distribution. The probability of his collection earning him the Langkawi trip is 0.323.

X~B(6 , 0.323). Calculate P(X=2) and P(X=6).

(15) X~P(1.2)

$\displaystyle P(X=m)=\frac{e^{-1.2}(1.2)^m}{m!}<0.03 $

You can try substituting positive integers for m=1,2,3,... till it satisfies the condition. In this case, m=4 since P(X=4)=0.026<0.03.

Part(2): In one day, $\displaystyle P(X=0)=e^{-1.2}=0.301$ but in n consecutive days, the probability is 0.301^n. Solve for 0.301^n>0.0002 by taking logs.