Question

Let $S$ be the sample space of the random experiment of throwing simultaneously two unbiased dice with six faces $\text{(numbered 1 to 6)}$ and $\text{EK= }\\!\\!\\{\\!\\!\text{ (a,b)}\in \text{S:ab=k }\\!\\!\\}\\!\\!\text{ }$ for $\text{k}\ge \text{1}$. If $\text{Pk=p(EK)}$ for $\text{k}\ge \text{1}$, then correct among the following is:
1. ${P_1} < {P_{30}} < {P_4} < {P_6}$
2. ${P_{36}} < {P_6} < {P_2} < {P_4}$
3. ${P_1} < {P_{11}} < {P_4} < {P_6}$
4. ${P_{36}} < {P_{11}} < {P_6} < {P_4}$

Answer 1

Firstly we will find out the total number of elements in sample space then we will calculate the probability of all the events ${{P}_{1,}},{{P}_{2}},{{P}_{3}}..........$ by finding the number of favourable outcomes. After that we will check given options one by one to find out which option is correct among them.

Complete step by step answer:
We know that in different situations the measure of uncertainty is called probability. The ratio of favourable number of outcomes to the total number of outcomes is the classical theory of probability .In statistical concept the probability is based on observations and collection of facts but in modern reference in axiomatic approach of probability we use some universal truth concepts.
The formula of the probability of an event is:
$\text{probability}=\dfrac{\text{number of desired outcomes}}{\text{total number of favourable outcomes}}$
Or
$P(A)=\dfrac{n(A)}{n(S)}$

Where, $P(A)$ is the probability of an event $A$
$n(A)$ is the number of favourable outcomes
And $n(S)$ is the total number of possible outcomes of a set.
If the probability of occurring an event is $P(A)$ then the probability of not occurring an event is $P(A')=1-PA$
Now, according to the given question:
The total number of elements in the sample space is 6\times 6=$$$$\text{36}
Let us calculate probability of ${{\text{P}}_{1}}$
Hence, ${{\text{P}}_{1}}=P({{E}_{1}})$
$\text{(a,b)}\in S$ such that $\text{ab=k}$
If $\text{k}\ge \text{1}$ then $ab\ge \text{1}$
The sample space $S$ will be:

& \text{S=}\left( \text{1, 1} \right)\text{, }\left( \text{1, 2} \right)\text{, }\left( \text{1, 3} \right)\text{, }\left( \text{1, 4} \right)\text{, }\left( \text{1, 5} \right)\text{ ,}\left( \text{1, 6} \right)\text{,} \\\ & \left( \text{2, 1} \right)\text{, }\left( \text{2, 2} \right)\text{,}\left( \text{2, 3} \right)\text{, }\left( \text{2, 4} \right)\text{, }\left( \text{2, 5} \right)\text{ ,}\left( \text{2, 6} \right)\text{,} \\\ & \left( \text{3, 1} \right)\text{, }\left( \text{3, 2} \right)\text{, }\left( \text{3, 3} \right)\text{, }\left( \text{3, 4} \right)\text{, }\left( \text{3, 5} \right)\text{ ,}\left( \text{3, 6} \right)\text{,} \\\ & \left( \text{4, 1} \right)\text{, }\left( \text{4, 2} \right)\text{, }\left( \text{4, 3} \right)\text{, }\left( \text{4, 4} \right)\text{, }\left( \text{4, 5} \right)\text{ ,}\left( \text{4, 6} \right)\text{,} \\\ & \left( \text{5, 1} \right)\text{, }\left( \text{5, 2} \right)\text{, }\left( \text{5, 3} \right)\text{, }\left( \text{5, 4} \right)\text{, }\left( \text{5, 5} \right)\text{ ,}\left( \text{5, 6} \right)\text{,} \\\ & \left( \text{6, 1} \right)\text{, }\left( \text{6, 2} \right)\text{, }\left( \text{6, 3} \right)\text{, }\left( \text{6, 4} \right)\text{, }\left( \text{6, 5} \right)\text{ ,}\left( \text{6, 6} \right) \\\ \end{aligned}$$ Hence the number of elements in sample space will be:$$\text{36}$$ $$\text{probability}=\dfrac{\text{number of desired outcomes}}{\text{total number of favourable outcomes}}$$ $${{\text{P}}_{1}}=\dfrac{36}{36}$$$$\text{=1}$$ Similarly, $${{\text{P}}_{2}}=P({{E}_{2}})$$ $$\text{(a,b)}\in S$$ such that $$\text{ab=k}$$ $$ab\ge 2$$ The sample space $$S$$ will be: $$\begin{aligned} & \text{S= }\left( \text{1, 2} \right)\text{, }\left( \text{1, 3} \right)\text{, }\left( \text{1, 4} \right)\text{, }\left( \text{1, 5} \right)\text{ ,}\left( \text{1, 6} \right)\text{,} \\\ & \left( \text{2, 1} \right)\text{, }\left( \text{2, 2} \right)\text{,}\left( \text{2, 3} \right)\text{, }\left( \text{2, 4} \right)\text{, }\left( \text{2, 5} \right)\text{ ,}\left( \text{2, 6} \right)\text{,} \\\ & \left( \text{3, 1} \right)\text{, }\left( \text{3, 2} \right)\text{, }\left( \text{3, 3} \right)\text{, }\left( \text{3, 4} \right)\text{, }\left( \text{3, 5} \right)\text{ ,}\left( \text{3, 6} \right)\text{,} \\\ & \left( \text{4, 1} \right)\text{, }\left( \text{4, 2} \right)\text{, }\left( \text{4, 3} \right)\text{, }\left( \text{4, 4} \right)\text{, }\left( \text{4, 5} \right)\text{ ,}\left( \text{4, 6} \right)\text{,} \\\ & \left( \text{5, 1} \right)\text{, }\left( \text{5, 2} \right)\text{, }\left( \text{5, 3} \right)\text{, }\left( \text{5, 4} \right)\text{, }\left( \text{5, 5} \right)\text{ ,}\left( \text{5, 6} \right)\text{,} \\\ & \left( \text{6, 1} \right)\text{, }\left( \text{6, 2} \right)\text{, }\left( \text{6, 3} \right)\text{, }\left( \text{6, 4} \right)\text{, }\left( \text{6, 5} \right)\text{ ,}\left( \text{6, 6} \right) \\\ \end{aligned}$$ Hence the number of elements in sample space will be:$$35$$ $$\text{probability}=\dfrac{\text{number of desired outcomes}}{\text{total number of favourable outcomes}}$$ $${{\text{P}}_{2}}=\dfrac{35}{36}$$ $${{\text{P}}_{3}}=P({{E}_{3}})$$ $$\text{(a,b)}\in S$$ such that $$\text{ab=k}$$ $$ab\ge 3$$ The sample space $$S$$ will be: $$\begin{aligned} & \text{S= }\left( \text{1, 3} \right)\text{, }\left( \text{1, 4} \right)\text{, }\left( \text{1, 5} \right)\text{ ,}\left( \text{1, 6} \right)\text{,} \\\ & \text{ }\left( \text{2, 2} \right)\text{,}\left( \text{2, 3} \right)\text{, }\left( \text{2, 4} \right)\text{, }\left( \text{2, 5} \right)\text{ ,}\left( \text{2, 6} \right)\text{,} \\\ & \left( \text{3, 1} \right)\text{, }\left( \text{3, 2} \right)\text{, }\left( \text{3, 3} \right)\text{, }\left( \text{3, 4} \right)\text{, }\left( \text{3, 5} \right)\text{ ,}\left( \text{3, 6} \right)\text{,} \\\ & \left( \text{4, 1} \right)\text{, }\left( \text{4, 2} \right)\text{, }\left( \text{4, 3} \right)\text{, }\left( \text{4, 4} \right)\text{, }\left( \text{4, 5} \right)\text{ ,}\left( \text{4, 6} \right)\text{,} \\\ & \left( \text{5, 1} \right)\text{, }\left( \text{5, 2} \right)\text{, }\left( \text{5, 3} \right)\text{, }\left( \text{5, 4} \right)\text{, }\left( \text{5, 5} \right)\text{ ,}\left( \text{5, 6} \right)\text{,} \\\ & \left( \text{6, 1} \right)\text{, }\left( \text{6, 2} \right)\text{, }\left( \text{6, 3} \right)\text{, }\left( \text{6, 4} \right)\text{, }\left( \text{6, 5} \right)\text{ ,}\left( \text{6, 6} \right) \\\ \end{aligned}$$ Hence the number of elements in sample space will be:$$33$$ $${{\text{P}}_{3}}=\dfrac{33}{36}$$ $$\text{probability}=\dfrac{\text{number of desired outcomes}}{\text{total number of favourable outcomes}}$$ Hence, we have find that probability of $${{P}_{1,}},{{P}_{2}},{{P}_{3}}$$ is so we can say that $${{P}_{1,}}>{{P}_{2}}>{{P}_{3}}..........>{{P}_{39}}>{{P}_{40}}$$ **So, the correct answer is “Option 4”.** **Note:** It should be noted that the definition of independent events is represented by the probability of the events, but the definition of mutually exclusive events is represented by the definition of events and no result in the mutually exclusive events is universal but in independent events the results may be universal.