Question

A determinant is chosen at random from the set of all determinants of order 2 with elements 0 or 1 only. The probability that the value of the determinant chosen is positive is k. Find 16k.

Answer 1

Here the given question is based on the concept of probability. We have to find the probability of the event choosing a set of all determinants of order 2. For this, first list out determinants of order 2 with element O or 1 only and in that list out determinants whose value is positive, then by using the definition of probability and on further simplification we get the required probability.

Complete step-by-step solution:
Probability is a measure of the likelihood of an event to occur. Many events cannot be predicted with total certainty. We can predict only the chance of an event to occur i.e., how likely they are to happen, using it. Probability can range in from 0 to 1, where 0 means the event to be an impossible one and 1 indicates a certain event.
The probability formula is defined as the probability of an event to happen is equal to the ratio of the number of favourable outcomes and the total number of outcomes.
$\text{Probability of event to happen}P\left( E \right) = \dfrac{\text{Number of favourable outcomes}}{\text{Total Number of outcomes}}$
Consider the given question:
A determinant is chosen at random from the set of all determinants of order 2 with elements 0 or 1 only.
Therefore, the total number of determinants $= {2^4} = 16$.
Let consider the event of the chose the value of determinant is positive:

1 & 1 \\\ 0 & 1 \\\ \end{matrix} \right|,\left| \begin{matrix} 1 & 0 \\\ 1 & 1 \\\ \end{matrix} \right|,\left| \begin{matrix} 1 & 0 \\\ 0 & 1 \\\ \end{matrix} \right| \right\\}$$ Therefore, total positive determinants $$ = 3$$ Given, k be the probability that the value of the determinant chosen is positive. By, the definition of probability $$ \Rightarrow \,\,\,\,P\left(\text{positive determinants} \right) = \dfrac{\text{Number of positive determinants}}{\text{Total Number of determinants}}$$ $$ \Rightarrow \,\,\,k = \dfrac{3}{{16}}$$ Now, find the value of $$16k$$ $$ \Rightarrow \,\,16k = 16 \times \dfrac{3}{{16}}$$ On simplification, we get $$\therefore \,\,16k = 3$$ Hence, the probability that the value of determinant chosen is positive is $$k = \dfrac{3}{{16}}$$ and **The value of $$16k = 3$$.** **Note:** Remember, In 2nd order determinant the determinant has 4 elements, so we can arrange 4 elements in $${2^4} = 16$$ ways. If the determinant value is positive the product of the element of 1st row 1st column with the element of 2nd row 2nd column should be greater than the product of element of 1st row 2nd column with element of 2nd row 1st column. The probability is a number of possible values that must know the definition. We can also find possible numbers by using the permutation concept or combination concept to make the problem easier.