A box contains 3 orange balls, 3 green balls and 2 blue balls. Three balls are drawn at random from the box without replacement. The probability of dr

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...

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)`

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)) =...

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 - B `P=15/25,q=10/25,n=10` `sigma^(2)=npq=10.""3/5.2/5=12/5`

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`