There are 12 balls in a bag, where n of them are white balls. If 2 balls are chosen at random from the bag without replacement, the probability of choosing at least 1 white balls is 19/33. Find n.
number of white --- n
number of non-white -- 12-n
prob of choosing at least 1 white
= 1 - prob(choosing both non-white)
= 1 - (12-n)/12 * (11-n)/11
= 19/33
(12-n)(11-n)/132 = 1 - 19/33 = 14/33
(132 - 23n + n^2)/12 = 14/3
3n^2 - 69n + 396 = 168
3n^2 - 69n +228 = 0
n^2 - 23n + 76 = 0
(n+19)(n-4) = 0
n = 4 or n = -19 , rejecting the negative number of balls
n = 4
check:
prob of picking 2 non-white
= (8/12)(7/11) = 56/132 = 14/33
prob of picking at least 1 white
= 1 - 14/33 = 19/33
Oh I see! I got the formula correct but I multiply it wrongly¡Thanks a lot!
To find the value of n (number of white balls), we can use the concept of probability.
First, let's consider the total number of ways to choose two balls from the 12 balls in the bag without replacement. This can be calculated using the combination formula (nCr), where n represents the total number of balls and r represents the number of balls we want to choose (in this case, r = 2):
C(12, 2) = 12! / (2! * (12-2)!)
Simplifying this expression, we get:
C(12, 2) = 12! / (2! * 10!) = (12 * 11) / (2 * 1) = 66
So, there are 66 different ways to choose two balls from the bag.
Now, let's focus on finding the number of ways to choose two balls without replacing them, where at least one of them is white. There are two scenarios where this can happen:
1) Choosing 2 white balls: The number of ways to choose 2 white balls from n white balls is given by C(n, 2).
2) Choosing 1 white ball and 1 non-white ball: The number of ways to choose 1 white ball from n white balls is given by C(n, 1), and the number of ways to choose 1 non-white ball from the remaining (12 - n) non-white balls is given by C(12 - n, 1). Multiplying these two values together will give us the total number of ways to choose 1 white ball and 1 non-white ball.
So, the total number of ways to choose two balls without replacement where at least one of them is white is:
C(n, 2) + C(n, 1) * C(12 - n, 1)
Now, we can set up an equation using the probability of choosing at least 1 white ball:
Probability = (C(n, 2) + C(n, 1) * C(12 - n, 1)) / 66
Given that the probability is 19/33, we can write:
19/33 = (C(n, 2) + C(n, 1) * C(12 - n, 1)) / 66
Solving this equation will give us the value of n, which represents the number of white balls.