2 ques.

[Sinistral]

in this ques what is the meaning of the 2nd statement: "A gambler observes the sequence of outcomes ............. first time that Success occurs". ??

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and please help me with the 2nd ques.

[Sinistral]

one more. i have problem with second part only (ie Q 10). actually by my calculations ans is coming out to be 2/3 which isnt an option. please someone help.

[duck]

Hi.. :)

H)Let say the gambler tosses a coin ( suppose, Head=success and Tail=failure). The given means that the gambler observes the sequence of outcome of tosses. Eg: TTTH.
So, in this case he'll get  a prize of (2)^4 as he is observing success for the first time on fourth toss.
Now, its easy to compute the expectation.


E) We need to find the probability of both patients getting infected by the end of the seond day.
Now, it can be possible in the following ways:
(i) Both of them got infection on the first day. Prob = 1/4*1/4
(ii) One of them contracted on the first day. Prob = 1/4*3/4
We'll have two such cases ( A on first day or B on first day) . So, effectively prob= 2*1/4*3/4
(iii)Both of them got infected on the second day. Prob= 1/4*1/4*3/4*3/4

Just add them up. You'll get 121/256.

10) Prob(A|B) = P[5≦x≦8] / P[3≦x≦8] = (21/50) / (37/50) = 21/37
Note: For P[5≦x≦8] , integrate [1/25(10-x)] from 5 to 8.
For P[3≦x≦8] , integrate {[(1/25)(x)] from 3 to 5 } + {[1/25(10-x)] from 5 to 8}

[Sumit]

@Duck: will u plz explain how you come up with probability of Both of them got infected on the second day. i.e 1/4*1/4*3/4*3/4...is it means probability of both of them not infected on 1st day multiplied by probability of both of them got infected on 2nd day.(i.e 3/4x3/4*1/4x1/4).

n why we use integration in 10th question..

[duck]

Hi Sumit.. :)

Yup, thats the correct reasoning for 1/4*1/4*3/4*3/4

10) Because its a continuous random variable. So, to find the probability we need to integrate the pdf.

[Sinistral]

Hello,
thank you so much.

please if u cud verify for (H): is it?


for (10) isnt X the no. of loaves of bread sold by the baker in a day. so can he sell for eg say 1.999999 loaves of bread? Why is the random variable continuous?
yes, by integrating I am getting the ans to be 21/37 option (b).

For calculating the value of k (Q9) I did sigma (summation) and I got the same ans 1/25 .

[duck]

Hi Sinistral.. :)

H) is correct.

10) X is not taking finite values.So, its not a discrete random variable.Hence, you cannot use summation.
It is taking values in intervals.So, must be a continuous random variable.

[Ayushya Kaul]

Could you show me how you calculated that integral? I don't get 3 as an answer!

[duck]

Hi Ayusha.. :)

"3" is not what you get from integration. Thats a different question.

[Kuldeep]

Hi sinistral,

Can. Pls elaborate ur calculation for H .... Hw did u gt 3/4^x .. In d answereh

[Sinistral]

just read the explanation provided by duck

if success = S and Failure = F then he will succeed if one of the following happens:
S, FS, FFS, FFFS ....
Respective probabilities: 3/4 , (1/4)(3/4), (1/4)^2 (3/4), (1/4)^3(3/4)....
respective prizes:          2^1 , 2^2      ,    2^3           ,      2^4  ........

P(Success) = P(S)+ P(FS)+ P(FFS)+ P(FFFS) .... (write it using sigma.)
Expected value of gamblers's prize = P(S)*2^1 + P(FS)*2^2 + P(FFS)*2^3 + P(FFFS)*2^4 .....
            write the above in the sigma form and u will get the desired result (given in earlier post). Its basically a GP , so it can be easily summed till infinity to get the desired answer.