A bag contains 16 red balls, 8 green balls and 6 blue balls. One ball is drawn at random. The probability that it is blue ball is
(a) 1/6
(b) 1/5
(c) 1/30
(d) 5/6
1 Answers
Talk Doctor Online in Bissoy App
A bag contains 6 red, 3 black and 6 white balls. A ball is selected at random from the bag, what is the probability that the selected ball is not red?
Correct option is: A) \(\frac{3}{5}\) There are 6 red, 3 black and 6 white balls. \(\therefore\) Total number of balls = 6 + 3 + 6 = 15. Number of balls which are not red =...
2 Answers 1 views
A bag contains 6 red marbles, 3 blue marbles and 7 green marbles. If a marble is randomly selected from the bag, probability that it is blue
Correct option is: A)\(\frac 3{16}\) There are 6 red marbles, 3 blue marbles and 7 green marbles in the bag. \(\therefore\) Total No of marbles in the bag = 6+3=7= 16 \(\therefore\) Probability that selected marble...
2 Answers 1 views
A bag contains 3 white and 2 black balls and another bag contains 2 white and 4 black balls. One bag is chosen at random. From teh selected bag, one b
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`
2 Answers 1 views
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
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...
2 Answers 3 views
Three bags contain a number of red and white balls as follows Bag I : 3 red balls, Bag II : 2 red balls and 1 white balls and Bag III : 3 white balls.
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....
2 Answers 2 views
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
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`
2 Answers 1 views
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
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)`
2 Answers 1 views
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
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`
2 Answers 14 views
A bag contains 3 red, 7 white, and 4 black balls. If three balls are drawn from the bag, then find the probability that all of them are of the same co
Correct Answer - `(10)/(91)` The required probability is `(.^(3)C_(3) + .^(7)C_(3) + .^(4)C_(3))/(.^(14)C_(3)) = (1 + 35 + 4)/(14 xx 13 xx 2) = (40)/(14 xx 26) = (10)/(91)`
2 Answers 1 views
A bag contains b blue balls and r red balls. If two balls are drawn at random, the probability drawing two red balls is five times the probability 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)) =...
2 Answers 1 views