A box contains 15 green and 10 yellow balls. If 10 balls are randomly drawn, one-nyne, with replacement, then the variance of the number of green balls drawn is
A. `6/25`
B. `12/5`
C. 6
D. 4
Probability of getting one red (R ) ball=`P_(R)cdotP_(barR)cdotP_(barR)+P_(barR)cdotP_(R)cdotP_(barR)+P_(barR)cdotP_(baR)cdotP_(R)`
`=5/8cdot3/72/6+3/8cdot5/7cdot2/6+3/8cdot5/7cdot2/6+3/8cdot2/7cdot5/6`
`15/(4cdot7cdot6)+15/(4cdot7cdot6)+15/(4cdot7cdot6)`
`=4/56+5/56+5/56=15/56`
Probability of drawing 2 green balls and one blue ball
`=P_(G)cdotP_(G)cdotP_(B)+P_(B)cdotP_(G)cdotP_(G)+P_(G)cdotP_(B)cdotP_(G)`
`=3/8cdot2/7cdot2/6+2/8cdot3/7cdot2/6+3/8cdot2/7cdot2/6`
`=1/28+1/28+1/28=3/28`
Procedure of drawing th balls has to end at the rth draw. So, exactly two black balls are drawn in first (r-1) draws and third black ball is drawn in...
Correct Answer - C
In any trial P (getting white ball) = P (W) =1/2
P(getting black ball)`=P(B)=1//2`
Required event will occur if in the first six trial, 3 white balls...
Correct Answer - A
P (required)
= P (all are white) + P (all are red) + P (all are black)
`=1/6xx2/9xx3/12+3/6xx3/9xx4/12+2/6xx4/9xx2/12=6/648+36/648+40/648=82/648`
Correct Answer - B
`(b)` Let `A` be the event of drawing a black ball.
`A_(1)`, the event of choosing the first box.
`A_(2)`, the event of choosing the second box....