There are 6 red 8 blue 7 green in a bag If 5 are drawn with replacement what is probability

total balls 6+8+7=21
5 are drawn at a time
at least 3 red balls must be there out of 5 balls
so the there must be 3 or 4 or 5 red balls
for 3 red balls-( 6/21 *6/21*6/21*15/21*15/21) here 15/21 bcoz we have to choose 2 balls out of 15 rest balls
like this if we proceed for 4 balls and 5 balls and at last add them then we got the result.
9 years agoHelpfull: Yes(9) No(0)
total balls 6+8+7=21
5 are drawn at a time
at least 3 red balls must be there out of 5 balls
so the there must be 3 or 4 or 5 red balls
for 3 red balls-( 6/21 *6/21*6/21*15/21*15/21) here 15/21 bcoz we have to choose 2 balls out of 15 rest balls
like this if we proceed for 4 balls and 5 balls and at last add them then we got the result.
i.e. nothing but.. (6c3*15c2)/21c5 + (6c4 * 15c1)/21c5 + (6c5 * 15c0)/21c5
=2331/20349
which is 37/323
9 years agoHelpfull: Yes(7) No(0)
2331/20349
which is 37/323
9 years agoHelpfull: Yes(4) No(0)
for 3 red balls-(5c3)*( 6/21 *6/21*6/21*15/21*15/21)
for 4 red balls-(5c4)*( 6/21 *6/21*6/21*6/21*15/21)
for 5 red balls-(5c5)*( 6/21 *6/21*6/21*6/21*6/21)

add them and we got the result
9 years agoHelpfull: Yes(3) No(0)
atlest 3red
3red 4red 5red
(6/21)^3*(15/21)^2+(6/21)^4*(15/21)+(6/21)^5
9 years agoHelpfull: Yes(1) No(0)
259/2261
P(at least three red) =
((6c3*15c2)/21c5 + (6c4 * 15c1)/21c5 + (6c5 * 15c0)/21c5)
9 years agoHelpfull: Yes(0) No(2)
5c3*(2/7)^3 *(5/7)^2 + 5c4*(2/7)^4 *(5/7) + 5c5*(2/7)^5
9 years agoHelpfull: Yes(0) No(0)

There are 6 red, 8 blue, 7 green in a bag. If 5 are drawn with replacement what is probability at least 3 are red