Question

Two dice, one blue and one green are thrown at the same time. What is the probability that the sum of two numbers appearing on the top is dice is
(1) 9
(2) Greater than 10
(3) Less than or equal to 11

(a) $\left( 3 \right)\to \dfrac{35}{36}$
(b) $\left( 1 \right)\to \dfrac{1}{9}$
(c) $\left( 2 \right)\to \dfrac{1}{12}$
(d) $\left( 3 \right)\to \dfrac{31}{36}$

Answer 1

We solve this problem by using the probability formula that is
$\Rightarrow P=\dfrac{\text{number of possible outcomes }}{\text{total number of outcomes}}$
For finding the total number of ways we use the condition that if a dice is rolled $'n'$ or $'n'$ dice are rolled at a time then the total number of ways is given as ${{6}^{n}}$ because each dice has 6 possibilities.
Then we take all the possibilities such that we get the sum of numbers on two dice as 9, greater than 10 and less than or equal to 11 to find the probability of getting the required sum.
We also use other formula of probability that is
$\Rightarrow P\left( {\bar{E}} \right)=1-P\left( E \right)$
Where $\bar{E}$ is the event of not occurring the event $E$

Complete step by step answer:
We are given that two dice one is blue and other is green are thrown at the same time
We know that there will be 6 numbers from 1 to 6 on a dice.
Let us assume that the total number of ways of getting two numbers on two dice as $'N'$
We know that if a dice is rolled $'n'$ or $'n'$ dice are rolled at a time then the total number of ways is given as ${{6}^{n}}$ because each dice has 6 possibilities.
By using the above condition we get

& \Rightarrow N={{6}^{2}} \\\ & \Rightarrow N=36 \\\ \end{aligned}$$ Now, let us solve the three parts one by one. (1) Probability of getting the sum as 9 Now, let us take all the possibilities of getting the sum on two dice as 9 as follows (i) Getting 4 on blue and 5 on green (ii) Getting 5 on blue and 4 on green (iii) Getting 3 on blue and 6 on green (iv) Getting 6 on blue and 3 on green Here, we can see that there are a total of 4 possible outcomes such that the sum will be 9. Let us assume that the probability of getting sum as 9 as $${{P}_{1}}$$ We know that the formula of probability is given as $$\Rightarrow P=\dfrac{\text{number of possible outcomes }}{\text{total number of outcomes}}$$ By using the above formula we get $$\begin{aligned} & \Rightarrow {{P}_{1}}=\dfrac{4}{36} \\\ & \Rightarrow {{P}_{1}}=\dfrac{1}{9} \\\ \end{aligned}$$ Therefore the answer for first part can be written as $$\left( 1 \right)\to \dfrac{1}{9}$$ (2) Getting the sum greater than 10 We know that the maximum sum we get on the two dice is 12. So, we can say that the possible sum greater than 10 will be 11 and 12. Now, let us take all possibilities where we get the sum as 11 and 12 (i) Getting 5 on blue and 6 on green (ii) Getting 6 on blue and 5 on green (iii) Getting 6 on blue and 6 on green Here we can see that there are a total of 3 possible outcomes. Let us assume that the probability of getting the sum greater than10 as $${{P}_{2}}$$ Now, by using the formula of probability we get $$\begin{aligned} & \Rightarrow {{P}_{2}}=\dfrac{3}{36} \\\ & \Rightarrow {{P}_{2}}=\dfrac{1}{12} \\\ \end{aligned}$$ Therefore, the answer for second part can be written as $$\left( 2 \right)\to \dfrac{1}{12}$$ (3) Getting the sum less than or equal to 11 First let us find the probability of getting the sum greater than 11 Let us take all the possibilities of getting the sum greater than 11 that is getting the sum as 12 (i) Getting 6 on blue and 6 on green Let us assume that the probability of getting sum greater than 11 as $$x$$ Now, by using the formula of probability we get $$\Rightarrow x=\dfrac{1}{36}$$ Let us assume that the probability of getting the sum less than or equal to 11 as $${{P}_{3}}$$ We know that the formula of probability that is $$\Rightarrow P\left( {\bar{E}} \right)=1-P\left( E \right)$$ Where $$\bar{E}$$ is the event of not occurring the event $$E$$ Here we can see that probability of getting the sum less than or equal to 11 is nothing but not getting the sum greater than 11 Now, by using the above formula we get $$\begin{aligned} & \Rightarrow {{P}_{3}}=1-x \\\ & \Rightarrow {{P}_{3}}=1-\dfrac{1}{36} \\\ & \Rightarrow {{P}_{3}}=\dfrac{35}{36} \\\ \end{aligned}$$ Therefore, the answer for the third part can be written as $$\left( 3 \right)\to \dfrac{35}{36}$$ **So, the correct answer is “Option a, b and c”.** **Note:** Students may make mistakes in taking the number of possible outcomes in the first part. We have the possible outcomes for getting the sum as 9 are (i) Getting 4 on blue and 5 on green (ii) Getting 5 on blue and 4 on green (iii) Getting 3 on blue and 6 on green (iv) Getting 6 on blue and 3 on green Here, students may miss the possibilities (ii) and (iv) Here, both the given dice are identical in nature but they are of different colour. So, we can have the vice versa relations in the two dice given. Here the possibility (ii) is the vice versa relation to possibility (i) both can occur because we are not mentioned in any specific order.