The given problem involves probability with conditional events and can be solved using the concept of conditional probability.
First, let's define the scenario:
Let's consider the events:
We need to find \(P(A \mid B)\), the probability that a red ball was transferred given that a black ball is drawn.
By Bayes' Theorem, \(P(A \mid B) = \frac{P(B \mid A) \cdot P(A)}{P(B)}\).
The probability of transferring a red ball from Bag I, \(P(A)\), is:
P(A) = \frac{\text{Number of red balls in Bag I}}{\text{Total number of balls in Bag I}} = \frac{3}{10}
Given that a red ball is transferred to Bag II, the composition of Bag II becomes:
The probability of drawing a black ball from Bag II in this case is:
P(B \mid A) = \frac{5}{10} = \frac{1}{2}
\(P(B)\) is the probability of drawing a black ball considering both possibilities (transferred ball is red, black, or white):
P(B) = P(B \mid A) \cdot P(A) + P(B \mid A') \cdot P(A')
For \(A'\):
P(A') = 1 - P(A) = \frac{7}{10}
If a black ball is transferred:
If a white ball is transferred:
P(A'\text{ and transfer black}) = \frac{4}{10} and P(A'\text{ and transfer white}) = \frac{3}{10}
So,
P(B) = \frac{1}{2} \cdot \frac{3}{10} + \frac{6}{11} \cdot \frac{2}{5} + \frac{5}{11} \cdot \frac{3}{10}
= \frac{3}{20} + \frac{12}{55} + \frac{15}{110} = \frac{51}{110}
P(A \mid B) = \frac{P(B \mid A) \cdot P(A)}{P(B)} = \frac{\frac{1}{2} \cdot \frac{3}{10}}{\frac{51}{110}}
= \frac{3}{20} \times \frac{110}{51} = \frac{5}{18}
Therefore, the probability that the transferred ball was red, given that a black ball was drawn from Bag II, is \(\frac{5}{18}\).
Let E→Ball drawn from Bag II is black.
ER→Bag I to Bag II red ball transferred.
EB→Bag I to Bag II black ball transferred.
EW→Bag I to Bag II white ball transferred.
\(P(\frac{E_R}{E})=\frac{P(\frac{E}{ER})⋅P(ER) }{ P(\frac{E}{ER})P(ER)+P(\frac{E}{EB})⋅P(EB)+P(\frac{E}{EW})P(EW)}\)
Here,
P(ER)=\(\frac{3}{10}\), P(EB)=\(\frac{4}{10}\), P(EW)=\(\frac{3}{10}\)
and
\(P(\frac{E}{E_R})=\frac{5}{10},P(\frac{E}{E_B})=\frac{5}{10},P(\frac{E}{E_W})=\frac{5}{10}\)
\(∴P(\frac{E_R}{E})=\frac{\frac{15}{100} }{\frac{ 15}{100}+\frac{24}{100}+\frac{15}{100}}\)
=\(\frac{15}{54}\)
\(=\frac{5}{18}\)
So, the correct option is (B): \(\frac{5}{18}\)
If probability of happening of an event is 57%, then probability of non-happening of the event is
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
The equivalent resistance between the points \(A\) and \(B\) in the given circuit is \[ \frac{x}{5}\,\Omega. \] Find the value of \(x\). 
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.