A determinant is chosen at random from the set of all determinants of order $2$ with elements $0$ or $1$ only. The probability that value of the determinant chosen is positive is
A. $\dfrac{3}{{16}}$
B. $\dfrac{3}{8}$
C. $\dfrac{1}{4}$
D. None of these
Answer
301.2k+ views
Hint: The question is based on the concept of probability. Here, we need to find the probability of the event choosing a determinant from the set of all determinants of order $2$.First step is to list out determinants of order $2$ with element $0$ or $1$ only and now find out those determinants whose value is positive, then by using the definition of probability we’ll get the required probability.
Formula Used:
Probability formula
$P(E) = \dfrac{\text{Number of favorable outcomes}}{\text{Number of possible outcomes}}$
Complete step by step solution:
As we know that,
Probability is a measure of how likely an event is to occur. Many events are impossible to predict with absolute certainty. We can only predict the chance of an event occurring, i.e. how likely it is to occur, using it.
Given that,
A determinant is chosen at random from the set of all determinants of order $2$ with elements $0$ or $1$ only.
$\therefore $ Total number of determinants $ = {2^4} = 16$
Number of Possible outcomes $ = 16$
Consider the following event in which the chosen determinant value is positive:
$k = \left\{ {\left| {\begin{array}{*{20}{c}}
1&1 \\
0&1
\end{array}} \right|,\left| {\begin{array}{*{20}{c}}
1&0 \\
1&1
\end{array}} \right|,\left| {\begin{array}{*{20}{c}}
1&0 \\
0&1
\end{array}} \right|} \right\}$
It implies that total positive event $ = 3$
Number of favorable outcomes $ = 3$
Using probability formula,
$P(\text{Positive determinant}) = \dfrac{\text{Number of favorable outcomes}}{\text{Number of possible outcomes}}$
$k = \dfrac{3}{{16}}$
Option ‘A’ is correct
Note: Students must know that in the second order determinant it has four elements. So, we can arrange those elements in ${2^4} = 16$ ways. Also, If the determinant value is positive, the product of the elements of the first row first column with the elements of the second row second column should be greater than the product of the elements of the first row second column with the elements of the second row first column. Here, we can also solve the problem by using the concept of combination and permutation to make the problem easier.
Formula Used:
Probability formula
$P(E) = \dfrac{\text{Number of favorable outcomes}}{\text{Number of possible outcomes}}$
Complete step by step solution:
As we know that,
Probability is a measure of how likely an event is to occur. Many events are impossible to predict with absolute certainty. We can only predict the chance of an event occurring, i.e. how likely it is to occur, using it.
Given that,
A determinant is chosen at random from the set of all determinants of order $2$ with elements $0$ or $1$ only.
$\therefore $ Total number of determinants $ = {2^4} = 16$
Number of Possible outcomes $ = 16$
Consider the following event in which the chosen determinant value is positive:
$k = \left\{ {\left| {\begin{array}{*{20}{c}}
1&1 \\
0&1
\end{array}} \right|,\left| {\begin{array}{*{20}{c}}
1&0 \\
1&1
\end{array}} \right|,\left| {\begin{array}{*{20}{c}}
1&0 \\
0&1
\end{array}} \right|} \right\}$
It implies that total positive event $ = 3$
Number of favorable outcomes $ = 3$
Using probability formula,
$P(\text{Positive determinant}) = \dfrac{\text{Number of favorable outcomes}}{\text{Number of possible outcomes}}$
$k = \dfrac{3}{{16}}$
Option ‘A’ is correct
Note: Students must know that in the second order determinant it has four elements. So, we can arrange those elements in ${2^4} = 16$ ways. Also, If the determinant value is positive, the product of the elements of the first row first column with the elements of the second row second column should be greater than the product of the elements of the first row second column with the elements of the second row first column. Here, we can also solve the problem by using the concept of combination and permutation to make the problem easier.
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