A box has 5 blue and 4 red balls. One ball is drawn at random and replaced. Its colour is also not noted. Then, another ball is drawn at random. What is the probability of second ball being blue?
A box = {5 blue, 4 red}
Let `E_(1)` is the event that first ball drawn is blue, `E_(2)` is the event that first ball drawn is red and E is the event that second ball drawn is blue.
`therefore P(E )=P(E_(1))cdotP(E//E_(1))+P(E_(2))cdotP(E//E_(2))`
`=5/9cdot4/8+4/9cdot5/8=20/72+20/72=40/72=5/9`
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`
Let `E_(1)`=Event that first ball being red
and `E_(2)`= Event that exactly tw of the three balls being red
`therefore P_(E_(1))=P_(R)cdotP_(R)cdotP_(R)+P_(R)cdotP_(R)cdotP_(barR)+P_(R)cdotP_(barR)cdotP_(R)+P_(R)cdotP_(barR)cdotP_(barR)`
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`
Correct Answer - A
The total number of ways of making the second draw is `.^(10)C_(5)`.
The number of draws of 5 balls containing 2 balls common with first draw of...
Correct Answer - A::B
Let the number of red and blue balls be r and b, respectively. Then, the probability of drawing two red balls is
`p_(1) = (.^(r )C_(2))/(.^(r+b)C_(2)) =...
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`