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Q13:
An instructor has a question 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 question bank, what is the probability that it will be an easy question given that it is a multiple choice question?
Solution :
Step 1: Classification of Data and Sample Space Formulation
To determine the conditional probability, we first categorize the total number of questions based on their type (True/False or Multiple Choice) and difficulty (Easy or Difficult). Let the events be defined as:
- $E$: The selected question is an Easy question.
- $M$: The selected question is a Multiple Choice question.
| Category | Easy | Difficult | Total |
|---|---|---|---|
| True/False ($T$) | 300 | 200 | 500 |
| Multiple Choice ($M$) | 500 | 400 | 900 |
| Total | 800 | 600 | 1400 |
The total number of questions in the bank is $N = 300 + 200 + 500 + 400 = 1400$.
Step 2: Identifying Required Probabilities
We are tasked with finding the conditional probability $P(E|M)$, which is defined as the probability that a question is easy, given that it is a multiple choice question. By the definition of conditional probability: $P(E|M) = \frac{P(E \cap M)}{P(M)}$ [By the definition of Conditional Probability]
First, calculate $P(M)$, the probability of selecting a Multiple Choice question:
$P(M) = \frac{\text{Total number of Multiple Choice questions}}{\text{Total number of questions}}$ $P(M) = \frac{500 + 400}{1400} = \frac{900}{1400} = \frac{9}{14}$Next, calculate $P(E \cap M)$, the probability of selecting a question that is both Easy and a Multiple Choice question:
$P(E \cap M) = \frac{\text{Number of Easy Multiple Choice questions}}{\text{Total number of questions}}$ $P(E \cap M) = \frac{500}{1400} = \frac{5}{14}$Step 3: Calculating Conditional Probability
Substitute the values into the conditional probability formula:
$P(E|M) = \frac{\frac{500}{1400}}{\frac{900}{1400}}$ $P(E|M) = \frac{500}{900}$ $P(E|M) = \frac{5}{9}$Final Solution: The probability that an easy question is selected, given that it is a multiple choice question, is $\frac{5}{9}$.
More Questions from Class 12 Mathematics Probability EXERCISE 13.1
- Q1: Given that E and F are events such that $P(E) = 0.6$, $P(F) = 0.3$ and $P(E \cap F) = 0.2$, find $P(E|F)$ and $P(F|E)$
- Q2: Compute $P(A|B)$, if $P(B) = 0.5$ and $P(A \cap B) = 0.32$
- Q3(i): If $P(A) = 0.8$, $P(B) = 0.5$ and $P(B|A) = 0.4$, find (i) $P(A \cap B)$
- Q3(ii): If $P(A) = 0.8$, $P(B) = 0.5$ and $P(B|A) = 0.4$, find (ii) $P(A|B)$
- Q3(iii): If $P(A) = 0.8$, $P(B) = 0.5$ and $P(B|A) = 0.4$, find (iii) $P(A \cup B)$
- Q4: Evaluate $P(A \cup B)$, if $2P(A) = P(B) = \frac{5}{13}$ and $P(A|B) = \frac{2}{5}$
- Q5(i): If $P(A) = \frac{6}{11}$ , $P(B) = \frac{5}{11}$ and $P(A \cup B) = \frac{7}{11}$, find (i) $P(A\cap B)$
- Q5(ii): If $P(A) = \frac{6}{11}$ , $P(B) = \frac{5}{11}$ and $P(A \cup B) = \frac{7}{11}$, find (ii) $P(A|B)$
- Q5(iii): If $P(A) = \frac{6}{11}$ , $P(B) = \frac{5}{11}$ and $P(A \cup B) = \frac{7}{11}$, find (iii) $P(B|A)$
- Q6(i): Determine $P(E|F)$. A coin is tossed three times, where (i) E : head on third toss, F : heads on first two tosses
- Q6(ii): Determine $P(E|F)$. A coin is tossed three times, where (ii) E : at least two heads, F : at most two heads
- Q6(iii): Determine $P(E|F)$. A coin is tossed three times, where (iii) E : at most two tails, F : at least one tail
- Q7(i): Determine $P(E|F)$. Two coins are tossed once, where (i) E : tail appears on one coin, F : one coin shows head
- Q7(ii): Determine $P(E|F)$. Two coins are tossed once, where (ii) E : no tail appears, F : no head appears
- Q8: Determine $P(E|F)$. A die is thrown three times, E : 4 appears on the third toss, F : 6 and 5 appears respectively on first two tosses
- Q9: Determine $P(E|F)$. Mother, father and son line up at random for a family picture E : son on one end, F : father in middle
- Q10(a): A black and a red dice are rolled. (a) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.
- Q10(b): A black and a red dice are rolled. (b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.
- Q11(i): A fair die is rolled. Consider events $E = \{1,3,5\}$, $F = \{2,3\}$ and $G = \{2,3,4,5\}$. Find (i) $P(E|F)$ and $P(F|E)$
- Q11(ii): A fair die is rolled. Consider events $E = \{1,3,5\}$, $F = \{2,3\}$ and $G = \{2,3,4,5\}$. Find (ii) $P(E|G)$ and $P(G|E)$
- Q11(iii): A fair die is rolled. Consider events $E = \{1,3,5\}$, $F = \{2,3\}$ and $G = \{2,3,4,5\}$. Find (iii) $P((E \cup F)|G)$ and $P((E \cap F)|G)$
- Q12(i): Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that (i) the youngest is a girl,
- Q12(ii): Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that (ii) at least one is a girl?
- Q14: Given that the two numbers appearing on throwing two dice are different. Find the probability of the event ‘the sum of numbers on the dice is 4’.
- Q15: Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’.
- Q16: If $P(A) = \frac{1}{2}$, $P(B) = 0$, then $P(A|B)$ is
- Q17: If A and B are events such that $P(A|B) = P(B|A)$, then
CBSE Solutions for Class 12 Mathematics Probability
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