rvthryet wrote:
In the first case we consider (2,3),(2,5),(2,7) but why do we not consider (3,2), (5,2), (7,2). But then total no. of cases will be 6 and Probability cannot be 6/6 = 1. I m too confused
Help!!!
Yes, that is confusing. But consider this:
Scenario 1: P(sum odd)=Favorable outcomes/Total # of outcomes
Total # of outcomes=4C2=6, # of selection of 2 balls out of 4 balls.
Favorable outcome: two numbers can total odd if one number is even 2 and another odd 3,5,7. So two ball must be (2,3),(2,5),(2,7). So we have 3 favorable outcomes.
P(sum odd)=Favorable outcomes/Total # of outcomes=3/6=1/2
Favorable outcome is the PAIR of balls, therefore it doesn't matter it'll be (2,3) or (3,2), it's still ONE pair. On the other hand for the total # of outcomes we are also counting number of drawing of TWO balls out of 4.
BUT we can count this in another way (the way we did for scenario 2):
First ball 2, seond any odd =1/4*3/3(as after drawing the even ball there are 3 balls left and all are odd)=1/4
First ball any odd, second ball 2 =3/4*1/3(as after drawing the odd ball there are 3 balls left and only one is even)=1/4.
Final P=1/4+1/4=1/2
OR: 2*1/4*3/3=1/2
Scenario 2: Different situation. But again the odd sum can occur if either of drawing gives even 2 and another odd 3,5,7.
This can happen in TWO ways:
even (2), odd (3,5,7); OR
odd(3,5,7), even(2).
P(sum odd)=2(as we have 2 ways favorable sum can occur)*1/4(probability of picking even 2, 1 ball out of 4 balls)*3/4(probability of picking odd 3,5 or 7, 3 out of 4 balls)=2*1/4*3/4=3/8
We also can count the probability here with combination:
2*1C1*3C1/4C1*4C1=2*(1*3)/(4*4)=3/8
1C1=# of selections of even out of one
3C1=# of selections of odd out of three
4C1=TOTAL # of selections of first ball out of 4
4C1=TOTAL # of selections of second ball out of 4 (as after drawing the first one we put it back, so there are still 4 balls in the bag)
Multiplying this by 2 as there are two ways: ood/even or even/odd.
Hope it's clear.