This is a beautiful problem form ISI MStat 2017 PSB Problem 3, where we use the basics of Bijection principle and Vandermone's identity to solve this problem.
Consider an urn containing 5 red, 5 black, and 10 white balls. If balls are drawn without replacement from the urn, calculate the probability that in the first 7 draws, at least one ball of each color is drawn.
It may give you the intuition, there is atleast in the problem, so let's do complementing counting. Okay! I suggest you to travel that path to help you realize the complexity of that approach.
Nevertheless, let's algebrify the problem.
You want to select red balls,
blue balls, and
white balls so that
.
Now, ,
and
represents our desired scenario.
,
and
denotes total number of cases.
Now, for each such triplet () of the number of balls of each colour we have selected, we can select them in
.
Let .
Total Number of Ways we can select the 7 balls =
.
Total Number of Ways we can select the 7 balls such that at least one ball of each color is drawn =
Let, .
Observe that has a bijection with
.
.
Total Number of Ways we can select the 7 balls such that at least one ball of each color is drawn =
Stay Tuned! Stay Blessed!
This is a beautiful problem form ISI MStat 2017 PSB Problem 3, where we use the basics of Bijection principle and Vandermone's identity to solve this problem.
Consider an urn containing 5 red, 5 black, and 10 white balls. If balls are drawn without replacement from the urn, calculate the probability that in the first 7 draws, at least one ball of each color is drawn.
It may give you the intuition, there is atleast in the problem, so let's do complementing counting. Okay! I suggest you to travel that path to help you realize the complexity of that approach.
Nevertheless, let's algebrify the problem.
You want to select red balls,
blue balls, and
white balls so that
.
Now, ,
and
represents our desired scenario.
,
and
denotes total number of cases.
Now, for each such triplet () of the number of balls of each colour we have selected, we can select them in
.
Let .
Total Number of Ways we can select the 7 balls =
.
Total Number of Ways we can select the 7 balls such that at least one ball of each color is drawn =
Let, .
Observe that has a bijection with
.
.
Total Number of Ways we can select the 7 balls such that at least one ball of each color is drawn =
Stay Tuned! Stay Blessed!