Question

Consider the following events.
${{E}_{1}}$: Six fair dice are rolled and at least one die shows one six.
${{E}_{2}}$: Twelve fair dice are rolled and at least two dice show sixes.
Let ${{p}_{1}}$ is the probability of ${{E}_{1}}$ and ${{p}_{2}}$ is the probability of ${{E}_{2}}$. Which of the following is true?
a). ${{p}_{1}}$ > ${{p}_{2}}$
b). ${{p}_{1}}$= ${{p}_{2}}$ = 0.6651
c). ${{p}_{1}}$ < ${{p}_{2}}$
d). ${{p}_{1}}$ = ${{p}_{2}}$ = 0.3349

Answer 1

To solve the above question we will first find the probability of event ${{E}_{1}}$ and the probability of event ${{E}_{2}}$independently and then we will find the relationship among each of them. We will use the property that rolling of dice is an independent event for any number of dice rolled together.

Complete step-by-step solution:
We will start solving the question by dividing the question into parts of event ${{E}_{1}}$and event ${{E}_{2}}$.
${{E}_{1}}$ states that we are rolling six fair dice. And we know that there are 6 faces on each dice and on each dice 1, 2, 3, 4, 5, 6 are written respectively.
So, the probability that a dice shows six is $\dfrac{1}{6}$
And, the probability of occurring different digits other than 6 is $\dfrac{5}{6}$. So, we can say that the probability of not occurring 6 is $\dfrac{5}{6}$ for each fair dice.
So, we can say that probability of not occurring 6 in any of six dices rolled is ${{\left( \dfrac{5}{6} \right)}^{6}}$ because rolling of each dice is an independent event.
We know from the question that the probability of occurring at least one six when 6 fair dice are rolled is ${{p}_{1}}$.
So, we can say that ${{p}_{1}}$ = 1 – the probability that six has not occurred in any of the dice rolled.
Hence,${{p}_{1}}$ = 1 - ${{\left( \dfrac{5}{6} \right)}^{6}}$
Therefore, ${{p}_{1}}$ = 0.665
Now, we can see that in the event ${{E}_{2}}$ twelve fair dice are rolled and at least two dice show sixes.
So, in the mathematical term, we can write it as:
Probability of occurring at least one dice shows sixes when 12 fair dice are rolled = 1 – the probability that 6 has not occurred on any of the 12 dices – Probability that six have occurred on exactly one dice.
We know that probability of not occurring six for one dice = $\dfrac{5}{6}$.
So, the probability of not occurring six on any of 12 dice rolled = ${{\left( \dfrac{5}{6} \right)}^{12}}$, because each dice is rolled independently.
And, probability that six has occurred on exactly one dice = ${}^{12}{{C}_{1}}\times {{\left( \text{Probability of not occurring six one dice} \right)}^{11}}\times {{\left( \text{Probability of occuring six on one dice} \right)}^{1}}$
$={}^{12}{{C}_{1}}\times {{\left( \dfrac{5}{6} \right)}^{11}}\times {{\left( \dfrac{1}{6} \right)}^{1}}$ $=\dfrac{12\times {{\left( 5 \right)}^{11}}}{{{\left( 6 \right)}^{12}}}$= 0.269
We know from that question that ${{p}_{2}}$ is the probability of ${{E}_{2}}$.
So, ${{p}_{2}}$= 1 – the probability that 6 has not occurred on any of the 12 dices – Probability that six have occurred on exactly one dice.
Therefore, we can write ${{p}_{2}}=1-{{\left( \dfrac{5}{6} \right)}^{12}}-{}^{12}{{C}_{1}}\times {{\left( \dfrac{5}{6} \right)}^{11}}\times {{\left( \dfrac{1}{6} \right)}^{1}}$
Hence, probability of event${{E}_{2}}$ = ${{p}_{2}}=1-{{\left( \dfrac{5}{6} \right)}^{12}}-{}^{12}{{C}_{1}}\times {{\left( \dfrac{5}{6} \right)}^{11}}\times {{\left( \dfrac{1}{6} \right)}^{1}}$ = 0.618
Now, we can see that ${{p}_{1}}$ = 0.665 and ${{p}_{2}}$= 0.618. So, we can say that ${{p}_{1}}>{{p}_{2}}$ .
So, option (a) is our required answer.

Note: Students should not get confused with dependent and independent events. Rolling of multiple dice together, tossing of more than one coin at a time, etc. all are independent events. For example: Let us say that we have rolled two fair dice together and we are calculating the probability of occurring 6 one at least one dice. So, the occurrence of six on one of the dice does not affect or destroy the chances of occurrence of 6 on the other dice; both dice are independent of each other. That’s why we generally calculate the probability for each dice independently for any event and then multiply them to obtain the overall probability.