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Q8:
If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive? (Assume that the individual entries of the determinant are chosen independently, each value being assumed with probability $\frac{1}{2}$).
Solution :
Step 1: Establishing the Sample Space
A second-order determinant is represented as:
$D = \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix} = a_{11}a_{22} - a_{12}a_{21}$
Each entry $a_{ij}$ can take a value from the set $\{0, 1\}$. Since there are 4 positions and each position has 2 possible values, the total number of possible determinants is $2^4 = 16$. [Per the Fundamental Counting Principle].
Step 2: Defining the Condition for a Positive Determinant
We require the value of the determinant $D > 0$. Given the expression $D = a_{11}a_{22} - a_{12}a_{21}$, the condition $D > 0$ implies:
$a_{11}a_{22} - a_{12}a_{21} > 0 \implies a_{11}a_{22} > a_{12}a_{21}$
Step 3: Systematic Case Analysis
Since the entries are restricted to $\{0, 1\}$, the products $a_{11}a_{22}$ and $a_{12}a_{21}$ can only take values $0$ or $1$. For the inequality $a_{11}a_{22} > a_{12}a_{21}$ to hold, the only possibility is:
- $a_{11}a_{22} = 1$
- $a_{12}a_{21} = 0$
Step 4: Counting Favorable Outcomes
We analyze the two conditions independently:
Condition 1: $a_{11}a_{22} = 1$
This occurs only if $a_{11} = 1$ AND $a_{22} = 1$. There is $1$ way to satisfy this.
Condition 2: $a_{12}a_{21} = 0$
This occurs if $a_{12} = 0, a_{21} = 0$; OR $a_{12} = 0, a_{21} = 1$; OR $a_{12} = 1, a_{21} = 0$. There are $3$ ways to satisfy this.
By the Multiplication Rule for independent events, the total number of favorable outcomes is $1 \times 3 = 3$.
Step 5: Calculating Probability
The probability $P(E)$ is defined as the ratio of favorable outcomes to the total number of elements in the sample space. [By the Classical Definition of Probability].
$P(D > 0) = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{3}{16}$
Final Solution: The probability that the value of the determinant is positive is $\frac{3}{16}$.
More Questions from Class 12 Mathematics Probability Miscellaneous Exercise on Chapter 13
- Q1(i): $A$ and $B$ are two events such that $P(A) \neq 0$. Find $P(B|A)$, if (i) $A$ is a subset of $B$
- Q1(ii): $A$ and $B$ are two events such that $P(A) \neq 0$. Find $P(B|A)$, if (ii) $A \cap B = \phi$
- Q2(i): A couple has two children, (i) Find the probability that both children are males, if it is known that at least one of the children is male.
- Q2(ii): A couple has two children, (ii) Find the probability that both children are females, if it is known that the elder child is a female.
- Q3: Suppose that 5% of men and 0.25% of women have grey hair. A grey haired person is selected at random. What is the probability of this person being male? Assume that there are equal number of males and females.
- Q4: Suppose that 90% of people are right-handed. What is the probability that at most 6 of a random sample of 10 people are right-handed?
- Q5: If a leap year is selected at random, what is the chance that it will contain 53 tuesdays?
- Q6: Suppose we have four boxes A,B,C and D containing coloured marbles as given below: One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A?, box B?, box C?
- Q7: Assume that the chances of a patient having a heart attack is 40%. It is also assumed that a meditation and yoga course reduce the risk of heart attack by 30% and prescription of certain drug reduces its chances by 25%. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga?
- Q9(i): An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known: $P(A \text{ fails}) = 0.2$, $P(B \text{ fails alone}) = 0.15$, $P(A \text{ and } B \text{ fail}) = 0.15$. Evaluate the following probabilities (i) $P(A \text{ fails|B has failed})$
- Q9(ii): An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known: $P(A \text{ fails}) = 0.2$, $P(B \text{ fails alone}) = 0.15$, $P(A \text{ and } B \text{ fail}) = 0.15$. Evaluate the following probabilities (ii) $P(A \text{ fails alone})$
- Q10: Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.
- Q11: If A and B are two events such that $P(A) \neq 0$ and $P(B | A) = 1$, then
- Q12: If $P(A|B) > P(A)$, then which of the following is correct :
- Q13: If A and B are any two events such that $P(A) + P(B) – P(A \text{ and } B) = P(A)$, then
CBSE Solutions for Class 12 Mathematics Probability
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