An urn contains 14 white balls and 6 red balls. two balls are to be selected in succession, at random and without replacement what is the probability that i) both balls are white, ii) the first ball in white and the second ball is red, iii) one white ball and one red ball are selected, iv) would the probabilities change if the first ball after being identified is put back in the urn before the second ball is selected?
Let's calculate the probabilities step by step:
i) To find the probability that both balls are white, we need to calculate the probability of selecting one white ball first, and then another white ball without replacement.
The probability of selecting a white ball first is:
P(white first) = (number of white balls) / (total number of balls) = 14 / 20
After selecting a white ball, there will be 13 white balls remaining out of 19 balls in total. Therefore, the probability of selecting a second white ball without replacement is:
P(white second) = (number of remaining white balls) / (total number of remaining balls) = 13 / 19
To find the probability that both balls are white, we multiply the two probabilities:
P(both white) = P(white first) * P(white second) = (14/20) * (13/19) = 182/380 = 91/190
So, the probability that both balls are white is 91/190.
ii) To find the probability that the first ball is white and the second ball is red, we need to calculate the probability of selecting a white ball first and then a red ball.
The probability of selecting a white ball first is:
P(white first) = (number of white balls) / (total number of balls) = 14 / 20
After selecting a white ball, there will be 6 red balls remaining out of 19 balls in total. Therefore, the probability of selecting a red ball next, without replacement, is:
P(red second) = (number of red balls) / (total number of remaining balls) = 6 / 19
To find the probability that the first ball is white and the second ball is red, we multiply the two probabilities:
P(first white, second red) = P(white first) * P(red second) = (14/20) * (6/19) = 84/380 = 21/95
So, the probability that the first ball is white and the second ball is red is 21/95.
iii) To find the probability that one white ball and one red ball are selected, we need to calculate the probability of selecting a white ball and then a red ball, or the probability of selecting a red ball and then a white ball.
The probability of selecting a white ball first and then a red ball is given by:
P(white first, red second) = P(white first) * P(red second) = (14/20) * (6/19) = 21/95
The probability of selecting a red ball first and then a white ball is given by:
P(red first, white second) = P(red first) * P(white second) = (6/20) * (14/19) = 84/380 = 21/95
To find the probability that one white ball and one red ball are selected, we add these two probabilities:
P(one white and one red) = P(white first, red second) + P(red first, white second) = (21/95) + (21/95) = 42/95
So, the probability that one white ball and one red ball are selected is 42/95.
iv) If the first ball is put back in the urn before the second ball is selected, the probabilities will change.
i) The probability that both balls are white:
P(both white with replacement) = P(white first) * P(white second) = (14/20) * (14/20) = 196/400 = 49/100
ii) The probability that the first ball is white and the second ball is red:
P(first white, second red with replacement) = P(white first) * P(red second) = (14/20) * (6/20) = 84/400 = 21/100
iii) The probability that one white ball and one red ball are selected:
P(one white and one red with replacement) = P(white first, red second) + P(red first, white second) = (21/100) + (21/100) = 42/100 = 21/50
So, the probabilities change when the first ball is put back in the urn before the second ball is selected.
To calculate the probabilities, we need to determine the total number of possible outcomes and the number of favorable outcomes for each scenario.
i) Probability that both balls are white:
The total number of balls in the urn is 14 white + 6 red = 20 balls.
When the first ball is selected, there are 20 balls in total, and 14 of them are white.
Thus, the probability of selecting a white ball on the first draw is 14/20.
After the first white ball is removed, there are 19 balls left in the urn, and 13 of them are white.
So, the probability of selecting a white ball on the second draw, without replacement, is 13/19.
To calculate the probability of both balls being white, we multiply the probabilities of each draw:
P(both white) = (14/20) * (13/19) = 182/380 = 0.479
ii) Probability that the first ball is white and the second ball is red:
The probability of selecting a white ball on the first draw is 14/20, as explained above.
After removing the first white ball, there are 19 balls left, out of which 6 are red.
Thus, the probability of selecting a red ball on the second draw, without replacement, is 6/19.
To calculate the probability, we multiply the probabilities of each draw:
P(first white, second red) = (14/20) * (6/19) = 84/380 = 0.221
iii) Probability that one white ball and one red ball are selected:
Here, we can consider two different scenarios: white first, red second or red first, white second.
For the white first, red second scenario:
P(white first, red second) = (14/20) * (6/19) = 84/380 = 0.221
For the red first, white second scenario:
P(red first, white second) = (6/20) * (14/19) = 84/380 = 0.221
To get the total probability, we sum the probabilities of each scenario:
P(one white, one red) = P(white first, red second) + P(red first, white second) = 0.221 + 0.221 = 0.442
iv) Would the probabilities change if the first ball is put back in the urn before the second ball is selected?
If the first ball is put back into the urn after being identified, then the total number of balls remains the same for each draw.
Therefore, the probabilities for each scenario (both white, first white and second red, one white and one red) would remain the same.
If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.
1) 14/20 * (14-1)/(20-1) = ?
2) 14/20 * 6/(20-1) = ?
3) Same as 2.
4) Yes, the total amount of balls would remain the same.