A bag contains 5 black and 6 white balls; two balls are drawn at random. What is the probability that the balls drawn are white?
3/11
Understanding the Probability Problem
The question asks for the probability of drawing two white balls from a bag containing both black and white balls. Probability is defined as the ratio of the number of favorable outcomes to the total number of possible outcomes.
Total Number of Balls and Possible Outcomes
First, let's find the total number of balls in the bag:
- Number of black balls = 5
- Number of white balls = 6
- Total number of balls = 5 + 6 = 11
We are drawing two balls at random from these 11 balls. The total number of ways to choose 2 balls from 11 is given by the combination formula:
$\binom{n}{k} = \frac{n!}{k!(n-k)!}$
Here, $n=11$ (total balls) and $k=2$ (balls drawn). So, the total number of possible outcomes is:
$\text{Total outcomes} = \binom{11}{2} = \frac{11!}{2!(11-2)!} = \frac{11!}{2!9!} = \frac{11 \times 10 \times 9!}{ (2 \times 1) \times 9!} = \frac{11 \times 10}{2} = 11 \times 5 = 55$
There are 55 total ways to draw two balls from the bag.
Favorable Outcomes: Drawing Two White Balls
We want to find the probability that both balls drawn are white. We need to find the number of ways to choose 2 white balls from the 6 white balls available in the bag.
Using the combination formula again, with $n=6$ (white balls) and $k=2$ (white balls drawn):
$\text{Favorable outcomes} = \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6!}{2!4!} = \frac{6 \times 5 \times 4!}{ (2 \times 1) \times 4!} = \frac{6 \times 5}{2} = 3 \times 5 = 15$
There are 15 ways to draw two white balls from the bag.
Calculating the Probability
The probability of an event is the ratio of the number of favorable outcomes to the total number of possible outcomes.
$\text{Probability (Drawing two white balls)} = \frac{\text{Number of ways to draw two white balls}}{\text{Total number of ways to draw two balls}}$
$\text{Probability} = \frac{15}{55}$
This fraction can be simplified by dividing both the numerator and the denominator by their greatest common divisor, which is 5.
$\text{Probability} = \frac{15 \div 5}{55 \div 5} = \frac{3}{11}$
Summary of Calculation
| Description | Number | Calculation |
|---|---|---|
| Black balls | 5 | |
| White balls | 6 | |
| Total balls | 11 | 5 + 6 |
| Balls drawn | 2 | |
| Total ways to draw 2 balls | 55 | $\binom{11}{2}$ |
| Ways to draw 2 white balls | 15 | $\binom{6}{2}$ |
| Probability of drawing 2 white balls | 3/11 | $\frac{15}{55}$ |
The probability that the two balls drawn are white is $\frac{3}{11}$.
Important Questions from Probability
Two distinct natural numbers from 1 to 9 are picked at random. What is the probability that their product has 1 in its unit place?
Two dice are thrown. What is the probability that difference of numbers on them is 2 or 3 ?
Suppose that there is a chance for a newly constructed building to collapse, whether the design is faulty or not. The chance that the design is faulty is 10%. The chance that the building collapses is 95% if the design is faulty, otherwise it is 45%. If it is seen that the building has collapsed, then what is the probability that it is due to faulty design?
What is the probability that boys and girls sit alternatively?
What is the probability that P and Q take the two end positions?
