a box contains 5 red balls and 6 black balls in how many ways can 6 balls be selected so that there are atleast two balls of the same colour
To calculate the number of ways to select 6 balls from a box containing 5 red balls and 6 black balls, such that there are at least two balls of the same color, we can use a combinatorial approach.
Step 1: Calculate the total number of ways to select 6 balls from the box, disregarding the condition of having at least two balls of the same color.
This can be done using combinations. We have a total of 11 balls in the box, and we want to select 6 balls. The number of ways to do this is given by the formula:
C(n, r) = n! / (r! * (n - r)!)
Where n is the total number of items (balls) and r is the number of items (balls) we want to select.
In this case, we have:
C(11, 6) = 11! / (6! * (11 - 6)!)
= 11! / (6! * 5!)
Step 2: Calculate the number of ways to select 6 balls from the box such that all 6 balls are of the same color.
Since we have 5 red balls and 6 black balls, there are two possibilities: selecting only red balls or selecting only black balls. In each case, we need to select all 6 balls, which means we need to count the number of ways to select all 6 red balls and the number of ways to select all 6 black balls.
For selecting all 6 red balls, we only have 5 red balls in the box, so the number of ways to do this is simply 1.
For selecting all 6 black balls, we only have 6 black balls in the box, so the number of ways to do this is also 1.
Step 3: Calculate the number of ways to select 6 balls from the box such that there are at least two balls of the same color.
To find the number of ways to select 6 balls with at least two balls of the same color, we need to subtract the number of ways found in Step 2 from the total number of ways found in Step 1.
Total number of ways with at least two balls of the same color = Total number of ways - Number of ways with all 6 balls of the same color
Total number of ways with at least two balls of the same color = C(11, 6) - (number of ways with all 6 red balls + number of ways with all 6 black balls)
Total number of ways with at least two balls of the same color = C(11, 6) - (1 + 1)
Total number of ways with at least two balls of the same color = C(11, 6) - 2
Now, you can calculate the total number of ways to select 6 balls from the box such that there are at least two balls of the same color by finding the value of C(11, 6) - 2 using the formula mentioned in Step 1.
To determine the number of ways 6 balls can be selected from the given box such that there are at least two balls of the same color, we need to consider the following possibilities:
1. Selecting all red balls: There is only one way to select all the red balls since there are exactly 5 red balls in the box.
2. Selecting all black balls: Similarly, there is only one way to select all the black balls since there are exactly 6 black balls in the box.
3. Selecting 5 black balls and 1 red ball: There are 6 ways to select one red ball from the given box, and the remaining 5 balls must be black. Therefore, there are 6 ways to select 5 black balls and 1 red ball.
4. Selecting 5 red balls and 1 black ball: Similar to the previous case, there are 5 ways to select one black ball from the given box, and the remaining 5 balls must be red. Hence, there are 5 ways to select 5 red balls and 1 black ball.
5. Selecting 4 red balls and 2 black balls: To calculate this, we need to consider the combinations of selecting 4 red balls out of the 5 available red balls and 2 black balls out of the 6 available black balls. The formula for combinations is nCr = n! / ((n-r)! * r!). Therefore, there are 5C4 (5 choose 4) ways to select 4 red balls and 6C2 (6 choose 2) ways to select 2 black balls. This results in (5! / (1!*4!)) * (6! / (4!*2!)) = 5 * 15 = 75 ways to select 4 red balls and 2 black balls.
6. Selecting 4 black balls and 2 red balls: Similarly to the previous case, there are 6C4 (6 choose 4) ways to select 4 black balls and 5C2 (5 choose 2) ways to select 2 red balls. This results in (6! / (2!*4!)) * (5! / (3!*2!)) = 15 * 10 = 150 ways to select 4 black balls and 2 red balls.
To get the total number of ways to select 6 balls such that there are at least two balls of the same color, you just need to sum up the above possibilities:
Total number of ways = 1 (selecting all red) + 1 (selecting all black) + 6 (selecting 5 black and 1 red) + 5 (selecting 5 red and 1 black) + 75 (selecting 4 red and 2 black) + 150 (selecting 4 black and 2 red)
= 1 + 1 + 6 + 5 + 75 + 150
= 238
Therefore, there are 238 ways to select 6 balls from the given box such that there are at least two balls of the same color.
number of selections with no restrictions
= C(11,6) = 462
We can't have: zero red, 1 red, zero black, 1 black
which would be
C(5,0)xC(6,6) + C(5,1)xC(6,5) + C(6,0)xC(5,5) + C(6,1)xC(5,4)
= 1x1 + 5x6 + 1x1 + 6x5
= 62
prob(your event) = (462 - 62)62/462 = 200/231
i) 2 red,4 black => C(5,2)*C(6,4) =150
ii)3 red,3 black => C(5,3)*C(6,3) =200
iii)4red,2 black => C(5,4)*C(6,2) =75
Therefore total number of ways =
150+200+75 = 425 ways
Hope this helps! Cheers! :)