Q There are 6 boll in bag out of them 4 green and 2 red randomly picked two boll one

probality of selecting 1 red + probality of selecting 2 red =((2c1*4c1)/6c2)+((2c2)/(6c2))=3/5
10 years agoHelpfull: Yes(34) No(2)

PROBABILITY=1-BOTH TIME GREEN BALL
1 TIME GREEN BALL=4/6(4 OUT OF 5)
2 TME GREEN BALL=3/5(3 OUT OF 5)
1-(4/6*3/5)
=1-2/5=3/5

ANS=3/5
10 years agoHelpfull: Yes(27) No(2)
6 balls 4 green and 2 red
there are three cases in this question
I. first ball is red and second is not red
prob for this= (2/6)*(4/5)
II. first ball is not red but second ball is red
prob for this= (4/6)*(2/5)
III. if both the balls are red
prob for this= (2/6)*(1/5)
now,
required prob= prob(I+II+III)
= (4/15)*(4/15)*(1/15)
= 3/5 ans....
10 years agoHelpfull: Yes(11) No(0)
probability of no boll will red become
4c2/6c2=2/5
so probability of at list one is red is
1-2/5=3/5
10 years agoHelpfull: Yes(5) No(0)
(4c1*2c1+4c0*2c2)/6c2 = 9/15
10 years agoHelpfull: Yes(2) No(2)
2C1*4C1+2C2/6C2=3/5
10 years agoHelpfull: Yes(2) No(0)
2(4/6*2/5)+(2/6*1/5)=3/5
10 years agoHelpfull: Yes(1) No(0)
1/15 2C2 / 6C2 = 1/15
10 years agoHelpfull: Yes(0) No(7)
1st ballis red 2nd ball maybe of any colour so an

ans =(2c1*5c1)/6c2=2/3
9 years agoHelpfull: Yes(0) No(0)
Thare are 2 cases... 1. one of the picked balls is red,other is green 2.two of the picked ball is red,other is none.
so according to both
((2c1*4c1)+(2c2*4c0))/6c2
=3/5
9 years agoHelpfull: Yes(0) No(0)

Q. There are 6 boll in bag out of them 4 green and 2 red, randomly picked two boll one by one so find the probabilty atleast one picked boll will be red?