Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

Two unbiased dice are thrown. If the probability that the sum of the numbers obtained on the two dice is neither a multiple of 2 nor a multiple of 3 is \[\dfrac{1}{p}\], find the value of p.

seo-qna
Last updated date: 22nd Jul 2024
Total views: 450.9k
Views today: 6.50k
Answer
VerifiedVerified
450.9k+ views
Hint: Write the total possible outcome that might occur in throwing two unbiased dice. From that formulate the pairs which will give the sum 2, 3, 4, 6, 8, 9, 10, 12 which are the multiples of either 2 or 3. Find the probability of its occurrence which will be P (E). Then find P (not E) i.e. we need to find the pair that don’t sum up to some multiple of 2 or 3.

Complete step-by-step answer:
It is said that two unbiased dice are drawn. The total number of possible outcomes from one dice is 6.

Therefore, the possible outcome that can be obtained from two dice are \[6\times 6=36\].
Let total outcome be marked as n.

\[\therefore n={{6}^{2}}=36\].

The possible outcomes are listed below.

\[\left\{ \begin{align}
  & \left( 1,1 \right),\left( 1,2 \right),\left( 1,3 \right),\left( 1,4 \right),\left( 1,5 \right),\left( 1,6 \right), \\

 & \left( 2,1 \right),\left( 2,2 \right),\left( 2,3 \right),\left( 2,4 \right),\left( 2,5 \right),\left( 2,6 \right), \\

 & \left( 3,1 \right),\left( 3,2 \right),\left( 3,3 \right),\left( 3,4 \right),\left( 3,5 \right),\left( 3,6 \right), \\

 & \left( 4,1 \right),\left( 4,2 \right),\left( 4,3 \right),\left( 4,4 \right),\left( 4,5 \right),\left( 4,6 \right), \\

 & \left( 5,1 \right),\left( 5,2 \right),\left( 5,3 \right),\left( 5,4 \right),\left( 5,5 \right),\left( 5,6 \right), \\

 & \left( 6,1 \right),\left( 6,2 \right),\left( 6,3 \right),\left( 6,4 \right),\left( 6,5 \right),\left( 6,6 \right) \\

\end{align} \right\}\]

Let E be the event that the sum of the numbers obtained on the dice is neither a multiple of 2 nor a multiple of 3.

Let E’ be the event that sum of numbers obtained on the dice is either multiple of 2 or 3,
i.e. the sum can vary as 2, 3, 4, 6, 8, 9, 10 and 12,

where 2, 4, 6, 8, 10 and 12 are multiples of 2, and 3, 6, 9, 12 are multiples of 3.

Here 12 is taken as the value of highest sum, as it is the sum of numbers obtained by adding the highest number in each dice. 6 is the highest number in both the dice.

Therefore, the sum = 6 + 6 = 12 is the highest sum for two unbiased dice.

Therefore, the sample space of the event E’ can be found from the total possible outcomes.

 \[\left\{ \begin{align}
  & \left( 1,1 \right),\left( 1,2 \right),\left( 1,3 \right),\left( 1,5 \right),\left( 2,1 \right),\left( 2,2 \right),\left( 2,4 \right),\left( 2,6 \right), \\

 & \left( 3,1 \right),\left( 3,3 \right),\left( 3,5 \right),\left( 3,6 \right),\left( 4,2 \right),\left( 4,4 \right),\left( 4,5 \right),\left( 4,6 \right), \\

 & \left( 5,1 \right),\left( 5,3 \right),\left( 5,4 \right),\left( 5,5 \right),\left( 6,2 \right),\left( 6,3 \right),\left( 6,4 \right),\left( 6,6 \right) \\

\end{align} \right\}\]

So there are 24 favorable possibilities out of 36 total possibilities.

\[\therefore n\left( E' \right)=24\]

Total possibilities, \[n\left( s \right)=36\].

Therefore, probability of an event E is given by P (E).

P (E) = number of simple events within THE / total number of possible outcomes.

\[P\left( E' \right)=\dfrac{n\left( E' \right)}{n\left( s \right)}=\dfrac{24}{36}\]

We know that \[P\left( E \right)=1-P\left( E' \right)\], where \[P\left( E' \right)\]can also be said as P (not E).

\[\therefore P\left( E \right)=1-\dfrac{24}{36}\]
                 \[=\dfrac{36-24}{36}=\dfrac{12}{36}=\dfrac{1}{3}\]

\[\therefore P\left( E \right)=\dfrac{1}{3}=\dfrac{1}{p}\]

From the question it was told that the probability is \[\dfrac{1}{p}\].

We got \[P\left( E \right)=\dfrac{1}{3}\].

Equate, \[\dfrac{1}{p}=\dfrac{1}{3}\].

Cross multiplying, we get p = 3.

Hence we got the value of p = 3.

Note: We found the number of pairs which give a sum that are multiples of either 2 or 3.
\[\therefore \] Pairs which don’t give their sum as a multiple of either 2 or 3 are 36 – 24 = 12.
So, we can directly make out \[P\left( E \right)=\dfrac{12}{36}=\dfrac{1}{3}\], and hence p = 3.