if there are 5 white balls and 3 red balls and 3 balls are drawn randomly frm them such the

at very first 3 balls are already drawn so there are 3 possibilities made 1)- 3 balls are drawn white so the prob of 4 ball drawn should be red will be 3c1/5c1 because 3 balls are already drawn so left are 5
2)- 2 balls drawn white and one red so 2c1/5c1
3)- 2 balls are drawn red and one is white so 1c1/5c1

so the prob becomes 3c1/5c1 * 2c1/5c1 * 1c1/5c1. = 6/125
5 years agoHelpfull: Yes(2) No(2)
First 3 balls are taken out. Make the cases of these three balls( just as in case of 3 coin flips). U will get 8 cases bbb, rrr etc. Now the case of rrrr will be discarded coz no red ball will be left for 4th turn. So take seven cases and insert red ball i.e. r in front of each case(bbbr , bbrr etc.) And find probability of all tgese cases without replacement and add them up . u ll get 0.375
8 years agoHelpfull: Yes(1) No(10)
how please explain @vivek
8 years agoHelpfull: Yes(0) No(2)
probability of picking red ball is 3C1/8C1=3/8
ans-0.375
8 years agoHelpfull: Yes(0) No(6)
The probability of picking red balls is:

WWWWW RRR -------------===8!/5!*3!

_ _ _ R _ _ _ _ ----------------- 7!/5!*2!

SO (7!/5!*2!) / (8!/5!*3!) ===3/8
4 years agoHelpfull: Yes(0) No(0)

if there are 5 white balls and 3 red balls.. and 3 balls are drawn randomly frm them such the the ball once taken is thrown away..
what is the probability that the 4th ball drawn is red.