Given three identical bags each containing 10 balls, whose colours are as follows:
| Bag I | 3 Red | 2 Blue | 5 Green |
| Bag II | 4 Red | 3 Blue | 3 Green |
| Bag III | 5 Red | 1 Blue | 4 Green |
A person chooses a bag at random and takes out a ball. If the ball is Red, the probability that it is from Bag I is $ p $ and if the ball is Green, the probability that it is from Bag III is $ q $, then the value of $ \frac{1}{p} + \frac{1}{q} $ is:
A gardener wanted to plant vegetables in his garden. Hence he bought 10 seeds of brinjal plant, 12 seeds of cabbage plant, and 8 seeds of radish plant. The shopkeeper assured him of germination probabilities of brinjal, cabbage, and radish to be 25%, 35%, and 40% respectively. But before he could plant the seeds, they got mixed up in the bag and he had to sow them randomly.
The center of a disk of radius $ r $ and mass $ m $ is attached to a spring of spring constant $ k $, inside a ring of radius $ R>r $ as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following Hooke’s law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disc, the time period of oscillation of center of mass of the disk is written as $ T = \frac{2\pi}{\omega} $. The correct expression for $ \omega $ is ( $ g $ is the acceleration due to gravity): 
Let $ a_0, a_1, ..., a_{23} $ be real numbers such that $$ \left(1 + \frac{2}{5}x \right)^{23} = \sum_{i=0}^{23} a_i x^i $$ for every real number $ x $. Let $ a_r $ be the largest among the numbers $ a_j $ for $ 0 \leq j \leq 23 $. Then the value of $ r $ is ________.
Conditional Probability is defined as the occurrence of any event which determines the probability of happening of the other events. Let us imagine a situation, a company allows two days’ holidays in a week apart from Sunday. If Saturday is considered as a holiday, then what would be the probability of Tuesday being considered a holiday as well? To find this out, we use the term Conditional Probability.
P(S | B) = P(B | B) = 1.
Proof of the same: P(S | B) = P(S ∩ B) ⁄ P(B) = P(B) ⁄ P(B) = 1.
[S ∩ B indicates the outcomes common in S and B equals the outcomes in B].
P(B | A), P(A) >0 or, P(A ∩ B) = P(B).P(A | B), P(B) > 0.
This theorem is named as the Multiplication Theorem of Probability.
Proof of the same: As we all know that P(B | A) = P(B ∩ A) / P(A), P(A) ≠ 0.
We can also say that P(B|A) = P(A ∩ B) ⁄ P(A) (as A ∩ B = B ∩ A).
So, P(A ∩ B) = P(A). P(B | A).
Similarly, P(A ∩ B) = P(B). P(A | B).
The interesting information regarding the Multiplication Theorem is that it can further be extended to more than two events and not just limited to the two events. So, one can also use this theorem to find out the conditional probability in terms of A, B, or C.
Read More: Types of Sets
Sometimes students get confused between Conditional Probability and Joint Probability. It is essential to know the differences between the two.