41 different balls are drawn from a bag containing balls labelled 1,2,…,300. The probability that the number of balls drawn that are labelled with an odd number is larger than the number of balls drawn that are labelled with an even number can be expressed as a/b where a and b are coprime positive integers. What is the value of a+b?
There are 150 odd balls and 150 even balls, the probability is the sum over the probabilities to draw k balls for k = 21 to 41. The equal number of even and odd balls implies that the probability that the number of even balls is larger is the same. Because there is no draw possible, both probabilities are 1/2.
an urn contains 6 balls identical in every aspect except color. theres is one yellow ball 2 red balls and 3 blue balls. you draw two balls from the urn but replace the first ball before drawing the second. find the probability that the first ball is blue and the second is red
To solve this problem, we need to consider the number of odd and even numbers in the bag and determine the probability of drawing more odd numbers than even numbers.
First, let's find the total number of odd and even numbers in the bag. Out of the numbers 1 to 300, there are:
- Odd numbers: 150 (1, 3, 5, ..., 299)
- Even numbers: 150 (2, 4, 6, ..., 300)
We want to draw 41 balls from the bag, so we need to determine the number of ways to pick more odd balls than even balls. To do this, we can consider each possibility of the number of odd balls drawn and sum them up.
Let's break it down:
- If we draw 0 odd balls, we must draw all 41 even balls. The probability of this happening is (150C0) * (150C41) / (300C41).
- If we draw 1 odd ball, we can choose any of the 150 odd balls to draw and select the remaining 40 even balls from the 150 even balls. The probability of this happening is (150C1) * (150C40) / (300C41).
- Similarly, we can calculate the probabilities for drawing 2, 3, ... odd balls.
By adding up the probabilities for drawing more odd balls than even balls, we get the desired probability.
Finally, we can solve the problem by computing the ratio a/b, where a is the sum of these probabilities and b is the denominator of the last calculated probability ratio a/b.
Let's calculate the value of a + b.