An instructor has a test bank consisting of 300 easy True/False questions, 200 difficult True/False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions.If a question is selected at random from the test bank, what is the probability that it will be an easy question given that it is a multiple choice question?
An instructor has a test bank consisting of 300 easy True/False questions, 200 difficult True/False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions.If a question is selected at random from the test bank, what is the probability that it will be an easy question given that it is a multiple choice question?
Solution and Explanation
Number of easy True/False questions = 300
Number of difficult True/False questions = 200
Number of easy multiple choice questions = 500
Number of difficult multiple choice questions = 400
Total number of all such questions = n(S) = 1400
Let,
E represents an easy question and F represents a multiple choice question.
\(∴n(E) = 300+500 = 800\)
And \(n(F) = 500+400 = 900\)
\(P(F)=\frac {n(F)}{n(S)}\)
\(P(F) =\frac {900}{1400}\)
\(n(E∩F)=500\)
⇒\(P(E∩F)=\frac {n(E∩F)}{n(S)}\)
\(P(E∩F) =\frac {500}{1400}\)
\(P(E|F)=\frac {P(E∩F)}{P(F)}\)
\(P(E|F)=\frac {500/1400}{900/1400}\)
\(P(E|F)=\frac {500}{900}\)
\(P(E|F)=\frac {5}{9}\)
Top Questions on Probability
- A bag contains 5 red, 3 blue, and 2 green balls. If two balls are drawn at random without replacement, what is the probability that both are red?
- BITSAT - 2025
- Mathematics
- Probability
A shop selling electronic items sells smartphones of only three reputed companies A, B, and C because chances of their manufacturing a defective smartphone are only 5%, 4%, and 2% respectively. In his inventory, he has 25% smartphones from company A, 35% smartphones from company B, and 40% smartphones from company C.
(ii) What is the probability that this defective smartphone was manufactured by company B?
A person buys a smartphone from this shopA shop selling electronic items sells smartphones of only three reputed companies A, B, and C because chances of their manufacturing a defective smartphone are only 5%, 4%, and 2% respectively. In his inventory, he has 25% smartphones from company A, 35% smartphones from company B, and 40% smartphones from company C.
A person buys a smartphone from this shop(i) Find the probability that it was defective.
- In a consignment of electric lamps, 5% are defective. If a random sample of 8 lamps are inspected, what is the probability that one or more lamps are defective?
- When a coin is tossed 4 times, what is the probability of getting at most 3 heads?
Questions Asked in CBSE CLASS XII exam
- What type of cell is mercury cell? Why is it more advantageous than dry cell?
- CBSE CLASS XII - 2025
- CBSE Compartment XII - 2025
- Electrochemistry
A compound (A) with molecular formula $C_4H_9I$ which is a primary alkyl halide, reacts with alcoholic KOH to give compound (B). Compound (B) reacts with HI to give (C) which is an isomer of (A). When (A) reacts with Na metal in the presence of dry ether, it gives a compound (D), C8H18, which is different from the compound formed when n-butyl iodide reacts with sodium. Write the structures of A, (B), (C) and (D) when (A) reacts with alcoholic KOH.
- CBSE CLASS XII - 2025
- Haloalkanes And Haloarenes
- State Faraday’s second law of electrolysis. How much electricity is required in terms of Faraday for the reduction of 1 mole of $Cr_2\text{O}_7^{2-} \text{ to Cr}^{3+}$?
- CBSE CLASS XII - 2025
- Electrochemistry
- Calculate the \(\Delta G\) of the following cell:
\[ \text{Mg}(s) | \text{Mg}^{2+} (aq) || \text{Cu}^{2+} (aq) | \text{Cu}(s) \] \[ \left[ E^\circ_{\text{Mg}^{2+}/\text{Mg}} = -2.37 \, \text{V}, E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34 \, \text{V}, F = 96500 \, \text{C mol}^{-1} \right] \]- CBSE CLASS XII - 2025
- Electrochemistry
- Calculate the emf of the following cell at 25°C : \[ \text{Zn(s)} | \text{Zn}^{2+} (0.1 \, \text{M}) \parallel \text{H}^+ (0.01 \, \text{M}) | \text{Hg}(l) | \text{Hg}_2^{2+} (1 \, \text{bar}), \, \text{Pt(s)} \] \[ \left[ E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \, \text{V}, \, E^\circ_{\text{H}_2/\text{H}^+} = 0.00 \, \text{V}, \, \log 10 = 1 \right] \]
- CBSE CLASS XII - 2025
- Electrochemistry
Concepts Used:
Conditional Probability
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.
Let’s discuss certain theorems of Conditional Probability:
- Let us consider a random experiment where the sample space S is considered as space and two events namely A and B happen there. Then, the formula would be:
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].
- Now let us consider any two events namely A and B happening in a sample space ‘s’, then, P(A ∩ B) = P(A).
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.