Question 971920: QNo1-A box contain 5 white and 7 black balls.two succesive drawn of 3 balls 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 these 3 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 6 other people .what is probability that there are 3 persons between them
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! With replacement, it is (5/12)(4/11)(3/10)=(60/1320) for white; for black,it is (7/12)*(6/11)*(5/10)=(210/1320). This product is [60*210]/[1320*1320]=(1/22)(21/132)=(1/22)(7/44)=7/968. Note: this is misleading, for with replacement means only replacing after the three white balls are drawn, which is not how it may be interpreted.
Without replacement, it is (5/12)(4/11)(3/10)(7/9)(6/8)(5/7)=(60/1320) (210/504)=(1/22)(210/504)
=(1/11)(105/504)=(1/11)(35/168)=35/1848
2. 60 and 65 = 0.7*0.4*0.8=0.224; 60 and 70 0.7*0.6*0.2=0.084; 65 and 70= 0.3*0.4*0.2=0.024; all 3 0.7*0.4*0.2=0.056. The sum is 0.338. The important thing to do here is to make sure you multiply by all three probabilities. The probability of two is straightforward, but if it isn't all of them, then you have to multiply by the probability the third person WON'T be part of it.
There are 8 people.
If x is in place 1 y has 1 places to be out of 7
x is in place 2, y has 1 places
etc. In every place x may stand, y can stand somewhere. The key is that it is THREE people, not at least 3 people.
This is 8/56=1/7
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