a bag contains 8 identical balls of which 5 are yellow and 3 are white.1 ball is removed,its colour is noted and is replaced in the bag.a second ball is drawn.by drawing a table,find the probability(a)both balls are yellow(b)both are white (c)only 1 is white.
Both yellow=(5/8)*(5/8)
Both white =(3/8)*(3/8)
one of each=2*(5/8)*(3/8)
should sum to 1.00
To find the probability of the different outcomes, we can create a table:
| Ball 1 | Ball 2 |
|--------|--------|
| Yellow | Yellow |
| Yellow | White |
| White | Yellow |
| White | White |
Now let's calculate the probabilities for each case:
(a) Both balls are yellow:
The probability of drawing a yellow ball on the first try is 5/8.
After replacing the ball, the probability of drawing another yellow ball is also 5/8.
So, the probability of both balls being yellow is (5/8) * (5/8) = 25/64.
(b) Both balls are white:
The probability of drawing a white ball on the first try is 3/8.
After replacing the ball, the probability of drawing another white ball is also 3/8.
So, the probability of both balls being white is (3/8) * (3/8) = 9/64.
(c) Only one ball is white:
There are two cases where only one ball is white:
1. Drawing a yellow ball on the first try and a white ball on the second try.
Probability = (5/8) * (3/8) = 15/64
2. Drawing a white ball on the first try and a yellow ball on the second try.
Probability = (3/8) * (5/8) = 15/64
Now, let's verify if the probabilities sum up to 1:
(25/64) + (9/64) + (15/64) + (15/64) = 64/64 = 1
Therefore, the probabilities calculated are correct.
To find the probabilities of the given events, you can use a table to keep track of the outcomes. Here's how you can do that:
Let's denote the event of drawing a yellow ball as "Y" and drawing a white ball as "W".
First, we need to find the probability of each event.
(a) Both balls are yellow:
We need to draw a yellow ball twice. Since the balls are replaced after each draw, the probability of drawing a yellow ball in the first draw is 5/8, and the same for the second draw. Therefore, the probability can be calculated as:
P(Y, Y) = (5/8) * (5/8)
(b) Both balls are white:
Similarly, we need to draw a white ball twice. The probability of drawing a white ball in the first draw is 3/8, and the same for the second draw. Therefore, the probability can be calculated as:
P(W, W) = (3/8) * (3/8)
(c) Only one is white:
To calculate this probability, we need to consider two cases: drawing a yellow ball first and then a white ball, or drawing a white ball first and then a yellow ball.
Case 1: Drawing a yellow ball first and then a white ball:
The probability of drawing a yellow ball first is 5/8, and the probability of drawing a white ball in the second draw is 3/8. Therefore, the probability of this case can be calculated as:
P(Y, W) = (5/8) * (3/8)
Case 2: Drawing a white ball first and then a yellow ball:
The probability of drawing a white ball first is 3/8, and the probability of drawing a yellow ball in the second draw is 5/8. Therefore, the probability of this case can be calculated as:
P(W, Y) = (3/8) * (5/8)
Now, to find the probability of only one ball being white, we need to sum the probabilities of both cases:
P(Only 1 white) = P(Y, W) + P(W, Y)
Make sure to simplify the fractions and calculate the probabilities accurately.
Finally, to verify that the probabilities sum up to 1, you can add the probabilities of all three events:
P(Y, Y) + P(W, W) + P(Only 1 white)
If the resulting sum is 1, then the probabilities have been calculated correctly.