Probability Calculation for Green Balls
This solution explains how to find the probability of randomly selecting two green balls from a box containing balls of different colors.
Total Balls Calculation
- Number of black balls = 2
- Number of green balls = 6
- Number of yellow balls = 4
- Total number of balls = $2 + 6 + 4 = 12$
Total Ways to Pick 2 Balls
We need to find the total number of ways to choose any 2 balls from the 12 available. This is a combination problem, denoted as $C(n, k)$ or $\binom{n}{k}$, where $n$ is the total number of items, and $k$ is the number of items to choose.
Total possible outcomes = $C(12, 2) = \frac{12!}{2!(12-2)!} = \frac{12!}{2!10!} = \frac{12 \times 11}{2 \times 1} = 66$
Favorable Ways to Pick 2 Green Balls
Next, we find the number of ways to choose exactly 2 green balls from the 6 green balls available.
Favorable outcomes = $C(6, 2) = \frac{6!}{2!(6-2)!} = \frac{6!}{2!4!} = \frac{6 \times 5}{2 \times 1} = 15$
Probability Calculation
The probability of an event is the ratio of the number of favorable outcomes to the total number of possible outcomes.
Probability (both balls are green) = $\frac{\text{Favorable outcomes}}{\text{Total possible outcomes}} = \frac{15}{66}$
To simplify the fraction, divide both the numerator and the denominator by their greatest common divisor, which is 3.
Probability (both balls are green) = $\frac{15 \div 3}{66 \div 3} = \frac{5}{22}$
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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?
