Question 972095: QNo1-A box contain 5 white and 7 black balls.two succesive drawn of 3balls are made 1)with replacement 2)without replacement .the probability that the first draw would produce white balls and the second draw would produce black balls are resp.
QNo-2.There are 3 persons aged 60,65,70 years old.the survival probabilities for these3 persons for another 5 years are 0.7,0.4 and 0.2 resp.what is probability that at least two of them would survive another five years
QNo-3.x and y stand in a line with 6other people.what is probability that thereare 3 persons between them
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Answers:i got the answers but the correct answers are
Q no2==0.388
Q no1==7/968 and 5/264
Q no3==1/7
pls provide me proper solution as the correct answers are given above....pls its urgent...send me solutions...
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! 12 balls
The only way I get the first one is to do the replacement AFTER the first three balls are drawn. I took the question to mean there is replacement AFTER each draw. The answer I got was double what is said. IF I replace after the first three balls, NOT what the question is asking, I get the second line below.
(5/12)^3 * (7/12)^3=125/1728 * 343/1728
(5/12)(4/11)(3/10) * (7/12)(6/11)(5.10)
(60/1320) (210/1320)=(1/22)(21/132)=(1/22)(7/44)=7/968.
WITH REPLACEMENT, AND TRUE REPLACEMENT
(5/12)(4/11)(3/10) *(7/9)(6/8)(5/7)
60/1320 * 210/504
1/22*105/252=(1/22)*(35/84)=(1/22)(5/12)=5/264
THE SOLUTION TO THE SURVIVAL ONE I POSTED ALREADY AND NEEDS TO BE CORRECTED. YOU NEED THE PROBABILITIES OF THE 60-65, 65-70,60-70 AND ALL THREE. The first is 0.7*0.4*0.8=0.224; second is 0.3*0.4*0.2=;0.024; third 0.7*0.6*0.2=0.084; all three 0.056=0.388
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