Question

a bag contains 6 yellow balls , 3 red balls and 2 green balls
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Answer 1

The correct question is:

A bag contains 6 yellow balls and 5 green balls. 4 balls are chosen at random from the bag. What is the probability of no yellow ball being chosen and at least one yellow ball being chosen?

 

Given:    

Total no of yellow balls = 6

Total no of green balls = 5

To find:

Now we have to find the probability of no yellow ball being chosen and also have to find the probability of at least one yellow ball being chosen.

   

Solution:

We all know that If we want to pick r number of items from n number of items, then the total number of ways in which you can do so is denoted by nCr. And nCr = n!r!×(n−r)! .

Let's call the event of no yellow ball that will choose (or only green balls which will choose) A, the event of at least one yellow ball that will choose B, and the sample space S. If A is to happen then, all of the balls have to be green which can happen in n(A) or ⁵C₄ ways. Since you are only picking four balls, so, n(S) = ¹¹C₄

Now calculate P(A) using the equation,

P(A) = n(A) / n(S)  

p(A) = ⁵C₄ / ¹¹C₄ = 5 / 330

Now, we need to understand that if A does not occur then, B will occur because not all the picked balls are going to be green.

Now, calculate P(B) using the the equation,

P(B) = 1−P(A)

p(B) = 1 - 5/330    

p(B) = 225/330  

So, the probability of no yellow ball that will choose is 5/330, and the probability of at least one yellow ball that will choose is 225/330

Answer 2

Answer:

The correct question is:

A bag contains 6 yellow balls and 5 green balls. 4 balls are chosen at random from the bag. What is the probability of no yellow ball being chosen and at least one yellow ball being chosen?

 

Given:    

Total no of yellow balls = 6

Total no of green balls = 5

To find:

Now we have to find the probability of no yellow ball being chosen and also have to find the probability of at least one yellow ball being chosen.

   

Solution:

We all know that If we want to pick r number of items from n number of items, then the total number of ways in which you can do so is denoted by nCr. And nCr = n!r!×(n−r)! .

Let's call the event of no yellow ball that will choose (or only green balls which will choose) A, the event of at least one yellow ball that will choose B, and the sample space S. If A is to happen then, all of the balls have to be green which can happen in n(A) or ⁵C₄ ways. Since you are only picking four balls, so, n(S) = ¹¹C₄

Now calculate P(A) using the equation,

P(A) = n(A) / n(S)  

p(A) = ⁵C₄ / ¹¹C₄ = 5 / 330

Now, we need to understand that if A does not occur then, B will occur because not all the picked balls are going to be green.

Now, calculate P(B) using the the equation,

P(B) = 1−P(A)

p(B) = 1 - 5/330    

p(B) = 225/330  

So, the probability of no yellow ball that will choose is 5/330, and the probability of at least one yellow ball that will choose is 225/330