A barket contain 3 red ball,5 blue ball and green ball.two ball are picked one after the other without replacement find the probability that
(a)both are red
(b)one it blue the other is red
(c)both are of different colour
(d)they are of the some colour
(e)the first is blue the other is from red
Mathematic
Answer quesion
Funny, you posted this twice, and still didn't say how many green balls there are. Let's say there were g of them.
That means there were N = 8+8 balls in total.
(a) 3/N * 2/(N-1)
(b) 2 * 5/N * 3/(N-1)
(c) ((3-1) + (5-1) + (g-1))/(N-1) = (N-3)/(N-1)
(d) 1 - (c)
(e) (b)/2
To find the probability of different events in this scenario, we need to determine the total number of balls in the basket and then calculate the number of favorable outcomes for each event.
Let's solve each part of the question step by step:
(a) Probability that both balls are red:
To calculate this probability, we need to find the number of favorable outcomes (picking two red balls) divided by the total number of possible outcomes.
Total balls in the basket: 3 red balls + 5 blue balls + 1 green ball = 9 balls
Favorable outcomes (both balls are red):
Selecting a red ball from 3 red balls = 3/9
For the second draw, since one red ball has been already taken, we have 2 red balls left out of 8 balls.
Selecting a second red ball from 2 red balls = 2/8
To find the probability of both balls being red, we multiply these probabilities:
P(both red) = (3/9) * (2/8) = 1/12
(b) Probability that one ball is blue and the other is red:
Again, the total balls in the basket remain the same: 3 red + 5 blue + 1 green = 9 balls.
Favorable outcomes (one blue and one red ball):
Selecting a blue ball from 5 blue balls = 5/9
For the second draw, there are 8 balls left, but now only 2 red balls are remaining.
Selecting a red ball from 2 red balls = 2/8
To find the probability of one blue ball and one red ball, we multiply these probabilities:
P(one blue and one red) = (5/9) * (2/8) = 5/36
(c) Probability that both balls are of different colors:
Total balls in the basket: 3 red + 5 blue + 1 green = 9 balls.
Favorable outcomes (both balls are different colors):
Selecting a red ball from 3 red balls = 3/9
For the second draw, we have 8 balls left, and now we can choose a blue or green ball from the remaining ones (5 blue and 1 green).
Selecting a blue or green ball from 6 balls = 6/8
To find the probability of both balls being of different colors, we multiply these probabilities:
P(both different colors) = (3/9) * (6/8) = 9/24 = 3/8
(d) Probability that they are of the same color:
Total balls in the basket: 3 red + 5 blue + 1 green = 9 balls.
Favorable outcomes (both balls are of the same color):
Selecting a red ball from 3 red balls = 3/9
For the second draw, we have 8 balls left, and we need to select another red ball from the remaining 2 red balls or a blue ball from the remaining 5 blue balls.
Selecting a red or blue ball from 7 balls = 7/8
To find the probability of both balls being of the same color, we multiply these probabilities:
P(both same color) = (3/9) * (7/8) = 21/72 = 7/24
(e) Probability that the first ball is blue and the other is from the red:
Total balls in the basket: 3 red + 5 blue + 1 green = 9 balls.
Favorable outcomes (first blue, second red):
Selecting a blue ball from 5 blue balls = 5/9
For the second draw, we have 8 balls left, and we need to select a red ball from the remaining 3 red balls.
Selecting a red ball from 3 red balls = 3/8
To find the probability of the first ball being blue and the second ball being red, we multiply these probabilities:
P(first blue, second red) = (5/9) * (3/8) = 15/72 = 5/24
Therefore, the probabilities are:
(a) Probability that both balls are red: 1/12
(b) Probability that one ball is blue and the other is red: 5/36
(c) Probability that both balls are of different colors: 3/8
(d) Probability that they are of the same color: 7/24
(e) Probability that the first ball is blue and the other is from the red: 5/24