A bag containing 5 red and 3 blue balls. If 3 balls are drawn at random without replacement the probability that exactly two of the three balls were r

Correct answer is (b) 1/5

Bag I ={3B,2W],Bag II={2B,4W} Let `E_(1)`=Event that bag I is selected `E_(2)`=Event that bag II is selected and E=Event that a black ball is selected `rArrP(E_(1))=1//2,P(E_(2))=1/2,P(E//E_(1))=3/5,P(E//E_(2))=2/6=1/3` `thereforeP(E)=P(E_(1))cdotP(E//E_(1))+P(E_(2))cdotP(E//E_(2))` `1/2cdot3/5+1/2cdot2/6=3/10+2/12` `=(18+10)/60=28/60=7/15`

Bag I : 3 red balls and 0 white ball Bag II : 2 red balls and 1 white ball. Bag III : 0 red ball andf 3 white balls....

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`

To draw 2 black, 4 white, and 3 red balls in order is same as arranging two black balls at first 2 places, 4 white balls at next 4 places,...

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)) =...

Number of ways of drawing 7 balls (second draw) = `.^(10)C_(7)` For each set of 7 balls of the second draw, 3 must be common to the set of 5...

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 - B The required probability is `(n^(2))/(""^(2n)C_(2))((n-1)^(2))/(""^(2n-2)C_(2))((n-2)^(2))/(""^(2n-4)C_(2))...(2^(2))/(""^(4)C_(2))(1^(2))/(""^(2)C_(2))` `=((1xx2xx3xx4xx...xx(n-1)n)^(2))/(((2n)!)/2^(n))=(2^(n)(n!)^(2))/((2n)!)=(2^(n))/(""^(2n)C_(n))`