Question 972093: 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
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
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Last problem
There are 8 people in line.
x can stand in place 1,2,3,4, each with 1/8 probability.
in place 1, y has 4 places out of 7
in place 2, y has 3 places out of 7
in place 3, y has 2 places out of 7
in place 4, y has 1 place out of 7
This doesn't give the answer, because I assumed at least. I assumed wrong.
For 3 people exactly, it is
place 1 (1/8) * (1/7) that y is in place 4
and place 2 the same for place 5
all the way to place 8 for x and place 5 for y
That is 8/56= 1/7
2.
60-65 2 SURVIVE, ONE doesn't. (0.7)(0.4)(0.8)=0.224
60-70 2 survive, one doesn't (0.7)(0.6)(0.2)=0.084
65-70 survive (0.3)(0.4)(0.2)=0.024
all survive (0.7)(0.4)(0.2)=0.056
Add them to get 0.388
For the balls.
with replacement it is (5/12)(4/11)(3/10) (7/12)(6/11)(5/10)=
60/1320 * 210/1320=(1/22) (7/44) =7/968
without 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)=35/1848
=5/264
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