an urn contains 6 white and 4 black balls .a fair die is rolled once and balls equal to the number appearing on the die is chosen from the urn,then the probability that the balls selected are only white ?
ans:1/5
my thought:let A be an event of choosing balls that are all whit and Bi be an event of choosing i balls("i" is the number obtained on the die)
then ,p(A)=SUMMATION i=1 to i=6( (P(Bi)*P(A/Bi))
let the number fallen on the die be 2, P(A)=P(B2)P(A/B2)=(1/6)(NO. OF WAYS CHOOSING 2 BALLS THAT IS WHITE/NO. OF WAYS OF CHOOSING 2 BALLS FROM 10 BALLS)
NOW ,HERE ,NO. OF WAYS CHOOSING 2 BALLS THAT IS WHITE IS 1 AND THAT IS WW
AND NO. OF WAYS CHOOSING 2 BALLS FROM WHOLE SET IS 3 AND THAT IS WW,WB,BB (W MEANS WHITE BALL ,B MEANS BLACK BALL)
BUT IN SOLUTION ITS IS GIVEN AS NO. OF WAYS CHOOSING 2 BALLS THAT IS WHITE IS 6C2 AND NO. OF WAYS CHOOSING 2 BALLS FROM SET IS 10C2 .....
MY QUESTION IS AS ALL BALLS ARE WHITE WHY DO WE NEED TO APPLY COMBINATIONS (LIKE 8C2),AS THERE IS ONLY ONE WAY OF CHOOSING BOTH WHITE BALLS ???
If the die shows $1$, there is a $6/10$ chance of choosing a white.
If the die shows $2$, there is a $6/10 \cdot 5/9$ chance of choosing two whites.
If the die shows $3$....
Can you continue?
Since each of these six die probabilities are $1/6$, multiply that by each case and sum.