A abg contains 5 red and blue balls. If 3 balls are drawn at random without replacement, them the probability of getting exactly one red ball is

Correct answer is (b) 1/5

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...

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`

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 - `1//4` If third ball is red, then in first two draws, there will be either no red ball or one red ball. Let events `R_(0)=` in first two...

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))`