Question

Find the probability that in 10 throws of a fair die a score which is a multiple of 3 will be obtained in at least 8 of the throws.

Answer 1

Hint: In order to solve this problem we have to know about some basic and interesting concepts in probability theory and also along with some concepts in permutations and combinations as well. Where the basic probability theorem is applied here, which is the probability of an event is given by the ratio of the number of favorable outcomes to the total number of possible outcomes of an event.

Complete step by step answer:
The probability of an event is given by the ratio of the number of favorable outcomes to the total number of outcomes in an event.
Given that in a fair play of dice, there are 10 throws in the fair die play.
We have to find out the probability of a score of a multiple of 3 in at least 8 of the throws.
Here in a die roll there are 6 outcomes, which are 1, 2, 3, 4, 5 and 6.
There is an equal probability for each of the outcomes as it is a fair die play.
The probability of each outcome is $P(X) = \dfrac{1}{6}$
Here the multiples of 3 are: 3 and 6
There are two multiples of 3, hence the probability of these numbers occurring in an event is given by:
$ \Rightarrow \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{2}{6}$
$ \Rightarrow \dfrac{1}{3}$ , is the success probability p.
$\therefore p = \dfrac{1}{3}$
Hence the failure probability is given by q :
$ \Rightarrow q = 1 - p$
$ \Rightarrow q = 1 - \dfrac{1}{3}$
$\therefore q = \dfrac{2}{3}$
We know that probability of r successes out of n possibilities is given by:
$ \Rightarrow P(X = r) = {}^n{c_r}{p^r}{q^{n - r}}$
$ \Rightarrow P(X = r) = {}^n{c_r}{p^r}{(1 - p)^{n - r}}$
As $q = 1 - p$.
So here there are 10 events, as there are 10 throws.
Now we have to calculate the probability of multiples of 3 in at least 8 throws, is given by:
$ \Rightarrow \sum\limits_{i = 8}^{10} {P(X = i)} $
$ \Rightarrow P(X = 8) + P(X = 9) + P(X = 10)$
\[ \Rightarrow {}^{10}{c_8}{\left( {\dfrac{1}{3}} \right)^8}{\left( {\dfrac{2}{3}} \right)^{10 - 8}} + {}^{10}{c_9}{\left( {\dfrac{1}{3}} \right)^9}{\left( {\dfrac{2}{3}} \right)^{10 - 9}} + {}^{10}{c_{10}}{\left( {\dfrac{1}{3}} \right)^{10}}{\left( {\dfrac{2}{3}} \right)^{10 - 10}}\]
Here we know that from permutations and combinations the formula of ${}^n{c_r} = \dfrac{{n!}}{{(n - r)!r!}}$, is used here.
\[ \Rightarrow \dfrac{{10!}}{{(10 - 2)!2!}}{\left( {\dfrac{1}{3}} \right)^8}{\left( {\dfrac{2}{3}} \right)^2} + \dfrac{{10!}}{{(10 - 1)!1!}}{\left( {\dfrac{1}{3}} \right)^9}{\left( {\dfrac{2}{3}} \right)^1} + \dfrac{{10!}}{{(10 - 0)!0!}}{\left( {\dfrac{1}{3}} \right)^{10}}{\left( {\dfrac{2}{3}} \right)^0}\]
\[ \Rightarrow \dfrac{{10!}}{{8!2!}}\left( {\dfrac{4}{{{3^{10}}}}} \right) + \dfrac{{10!}}{{9!1!}}\left( {\dfrac{2}{{{3^{10}}}}} \right) + \dfrac{{10!}}{{10!0!}}\left( {\dfrac{1}{{{3^{10}}}}} \right)\]
\[ \Rightarrow \dfrac{{10 \times 9}}{2}\left( {\dfrac{4}{{{3^{10}}}}} \right) + \dfrac{{10}}{1}\left( {\dfrac{2}{{{3^{10}}}}} \right) + \dfrac{1}{1}\left( {\dfrac{1}{{{3^{10}}}}} \right)\]
On further simplification and taking the common terms together as given below:
\[ \Rightarrow 90\left( {\dfrac{2}{{{3^{10}}}}} \right) + 10\left( {\dfrac{2}{{{3^{10}}}}} \right) + \left( {\dfrac{1}{{{3^{10}}}}} \right)\]
\[ \Rightarrow \dfrac{1}{{{3^{10}}}}\left( {90(2) + 10(2) + 1} \right)\]
\[ \Rightarrow \dfrac{{201}}{{{3^{10}}}} = 0.0034\]
$\therefore $The probability of a fair die a score in 10 throws, which is a multiple of 3 will be obtained in at least 8 of the throws is 0.0034

The probability is at least 8 throws of multiple of 3 in 10 throws is 0.0034.

Note: It is very important to note that here we are asked to find the probability of at least 8 throws out of 10 throws, here at least 8 throws means that 8 throws and more than 8 throws. That is, it is not just the 8th throw or 8th event, here we have to find the throws more than 8 including it, as it is given as at least 8 throws. This is the most striking mistake where most of us can go wrong.