Question
Question: Seven balls are drawn simultaneously from a bag containing 5 white and 6 green balls. The probabilit...
Seven balls are drawn simultaneously from a bag containing 5 white and 6 green balls. The probability of drawing 3 white and 4 green balls is:
1. 11C77
2. 11C7(5C3+6C4)
3. 11C7(5C2×6C2)
4. 11C7(6C3+5C4)
Solution
Let us name the two colours as colour 1 and colour 2. It is given that the number of balls from the bag which are drawn are 5 white and 6 green. Now we have to find the probability for drawing 3 white and 4 green balls. Let us assume that picking up a ball, the probability for this is equal to P. Now we will find the value of P. Now we will find the possibilities to pick the ball as per the question. We know that the probability to have exactly r trials among n trials is equal to P(X=r). Then, P(X=r)=nCrprqn−r, where q=1−p. Now by using this formula, we find the required probability.
Complete step-by-step solution:
According to the question, a total of seven balls are drawn simultaneously from a bag containing 5 white and 6 green balls and we are asked to find the probability of drawing 3 white and 4 green balls.
Since the total number of balls present in the bag are = 5 white + 6 green = 11 balls.
And we have to select 7 balls.
Therefore, total cases = 11C7.
So, favourable cases for white balls = 5C3.
And for green ball = 6C4.
So, clear favourable cases = 5C3.6C4=5C2.6C2. [nCr=nCn−r]
The required probability = 11C7(5C2×6C2).
Hence the correct answer is option 3.
Note: Students may have a misconception that if the probability to have exactly r trials among n trials is equal to P(X=r). Then, P(X=r)=nCrprqn−r, where q=1−p. But we know that if the probability to have exactly r trials among n trials is equal to P(X=r). Then, P(X=r)=nCrprqn−r, where q=1−p. OS, the misconception will lead to confusion. So, it should be avoided.