1me 2...2nd question not clear....is it saying dat wat is d probability dat it will last in d 2nd year??? ne1 help...
Q 2n girls are randomly divided into 2 subgroups contaitning n girls each.The probablity that two tallest girls are in diff groups...
Ans given is n2(2n-1)..but i am getting (n-1)/2n
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24Q2 the probablity that a certain component fails when first used is 0.1 .If i doesnt fail immediately,the probablity that it lasts for one yr is 0.99.Find prob that it will last for one yr ???
I didnt understand the ques at all....when prob that it will last one yr is given..what is the ques asking ??will it not be 0.99 ??
66Q1)No. of ways which favor the required event: First put the two tallest girls in the separate groups. This could be done in 1 way only. Now we need to divide the remaining 2n-2 girls in the two groups. To this end, select any n-1 girls from the 2n-2 girls and you are done with. So the required number is
2n-2Cn-1
On the other hand, the total number of ways of dividing the girls into two groups is 2nCn.
So the required probability
= 2n-2Cn-12nCn = n2(2n-1)
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106Q2 the probablity that a certain component fails when first used is 0.1 .If i doesnt fail immediately,the probablity that it lasts for one yr is 0.99.Find prob that it will last for one yr ???
The component will last for one year if it doesnt fail immediately and then it lasts for one year
So, prob = (1-0.1)*0.99 - 0.9*0.99 = 0.891
106Agar woh component first use par hi fail karta hai toh woh ek saal to last nahiin karega.
So, prob that it doesnt fail in first use = 0.9
Ab agar fail nahiin karta hai tab prob of lasting for one yr is 0.99.
Toh total prob = 0.9*0.99 (prob of not failing*prob of lasting one year)
agar yeh bhi samajh nahiin paaye then i cant xplain more
1can u put photos of it as may be arshad did for fiitjee paper help us