# Question 7 Exercise 6.4

Solutions of Question 7 of Exercise 6.4 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

## Question 7(i)

Two dice are thrown simultaneously. Find the probability of getting doublet of even numbers.

### Solution

The sample space rolling a pair of dice is \begin{align}S&=\{(i, j) ; i, j=1,2,3,4,5,6\}\\ &=\left[\begin{array}{llllll} (1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) \\ (2,1) & (2,2) & (2,3) & (2,4) & (2,5) & (2,6) \\ (3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6) \\ (4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6) \\ (5,1) & (5,2) & (5,3) & (5,4) & (5,5) & (5,6) \\ (6.1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6) \end{array}\right]\end{align} So total number of sample points are $$n(S)=6 \times 6=36$$

doublet of even numbers.

Let \begin{align}A&=\{(2,2),(4,4),(6,6)\}\\ n(A)&=3\end{align} Hence the possibility of getting doublet of an even number is: $$P(A)=\dfrac{n(A)}{n(S)}=\dfrac{3}{36}=\dfrac{1}{12}$$

## Question 7(ii)

Two dice are thrown simultaneously. Find the probability of getting a sum less than $6$.

### Solution

The sample space rolling a pair of dice is \begin{align} S&=\{(i, j) ; i, j=1,2,3,4,5,6\}\\ &=\left[\begin{array}{llllll} (1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) \\ (2,1) & (2,2) & (2,3) & (2,4) & (2,5) & (2,6) \\ (3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6) \\ (4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6) \\ (5,1) & (5,2) & (5,3) & (5,4) & (5,5) & (5,6) \\ (6.1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6) \end{array}\right]\end{align} So total number of sample points are $$n(S)=6 \times 6=36$$ A sum less than $6$

Let $B=\{$ a number less than 6$\}$,

then from sample space, we see that $n(B)=10$.

Thus the probability of getting a number less than 6 is: $$P(B)=\dfrac{n(B)}{n(S)}=\dfrac{10}{36}=\dfrac{5}{18}$$

## Question 7(iii)

Two dice are thrown simultaneously. Find the probability of getting a sum more than $7.$

### Solution

The sample space rolling a pair of dice is \begin{align} S&=\{(i, j) ; i, j=1,2,3,4,5,6\}\\ &=\left[\begin{array}{llllll} (1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) \\ (2,1) & (2,2) & (2,3) & (2,4) & (2,5) & (2,6) \\ (3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6) \\ (4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6) \\ (5,1) & (5,2) & (5,3) & (5,4) & (5,5) & (5,6) \\ (6.1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6) \end{array}\right]\end{align} So total number of sample points are $$n(S)=6 \times 6=36$$ A sum more than $7$

Let $C=\{$ a sum mure than 7$\}$,

then from sample space, we see that $n(C)=5$.

Thus the probability of getting number more than $7$ throwing dice two times is: $$P(C)=\dfrac{n(C)}{n(S)}=\dfrac{15}{36}=\dfrac{5}{12}$$

## Question 7(iv)

Two dice are thrown simultaneously. Find the probability of getting a sum greater than $10.$

### Solution

The sample space rolling a pair of dice is \begin{align} S&=\{(i, j) ; i, j=1,2,3,4,5,6\}\\ &=\left[\begin{array}{llllll} (1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) \\ (2,1) & (2,2) & (2,3) & (2,4) & (2,5) & (2,6) \\ (3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6) \\ (4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6) \\ (5,1) & (5,2) & (5,3) & (5,4) & (5,5) & (5,6) \\ (6.1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6) \end{array}\right]\end{align} So total number of sample points are $$n(S)=6 \times 6=36$$ A sum greater than $10$

Let $D=\{$ a sum greater than 10$\}$,

then from sample space, we get $n(D)=3$.

Thus the probability of getting number greater than $10$ is: $$P(D)=\dfrac{n(D)}{n(S)}=\dfrac{3}{36}=\dfrac{1}{12}$$

## Question 7(v)

Two dice are thrown simultaneously. Find the probability of getting a sum at least $10.$

### Solution

The sample space rolling a pair of dice is \begin{align}S=\{(i, j) ; i, j=1,2,3,4,5,6\}\\ &=\left[\begin{array}{llllll} (1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) \\ (2,1) & (2,2) & (2,3) & (2,4) & (2,5) & (2,6) \\ (3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6) \\ (4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6) \\ (5,1) & (5,2) & (5,3) & (5,4) & (5,5) & (5,6) \\ (6.1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6) \end{array}\right]\end{align} So total number of sample points are $$n(S)=6 \times 6=36$$ A sum at least $10$

Let $E=\{a$ sum at least 10$\}$,

then from sample space, we see that $n(E)=6$.

Thus the probability of getting at least $10$ is: $$P(E)=\dfrac{n(E)}{n(S)}=\dfrac{6}{36}=\dfrac{1}{6}$$

## Question 7(vi)

Two dice are thrown simultaneously. Find the probability of getting six as the product.

### Solution

The sample space rolling a pair of dice is \begin{align}S&=\{(i, j) ; i, j=1,2,3,4,5,6\}\\ &=\left[\begin{array}{llllll} (1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) \\ (2,1) & (2,2) & (2,3) & (2,4) & (2,5) & (2,6) \\ (3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6) \\ (4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6) \\ (5,1) & (5,2) & (5,3) & (5,4) & (5,5) & (5,6) \\ (6.1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6) \end{array}\right]\end{align} So total number of sample points are $$n(S)=6 \times 6=36$$ six as the product

Let $F=\{6$ as the product $\}$, then from sample space, we see that $n(F)=4$.

Thus the probability of getting numbers whose product is $6$ is: $$P(F)=\dfrac{n(F)}{n(S)}=\dfrac{4}{36}=\dfrac{1}{9}$$

## Question 7(vii)

Two dice are thrown simultaneously. Find the probability of getting an even number as the sum.

### Solution

The sample space rolling a pair of dice is \begin{align} S&=\{(i, j) ; i, j=1,2,3,4,5,6\}\\ &=\left[\begin{array}{llllll} (1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) \\ (2,1) & (2,2) & (2,3) & (2,4) & (2,5) & (2,6) \\ (3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6) \\ (4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6) \\ (5,1) & (5,2) & (5,3) & (5,4) & (5,5) & (5,6) \\ (6.1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6) \end{array}\right]\end{align} So total number of sample points are $$n(S)=6 \times 6=36$$ An even number as the sum

Let $G=\{$ an even number as sum $\}$,

then from sample space, we see that $n(G)=18$.

Thus the probability is: $$P(G)=\dfrac{P(G)}{n(S)}=\dfrac{18}{36}=\dfrac{1}{2}$$

## Question 7(viii)

Two dice are thrown simultaneously. Find the probability of getting an odd number as the sum.

### Solution

The sample space rolling a pair of dice is \begin{align}S&=\{(i, j) ; i, j=1,2,3,4,5,6\}\\ &=\left[\begin{array}{llllll} (1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) \\ (2,1) & (2,2) & (2,3) & (2,4) & (2,5) & (2,6) \\ (3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6) \\ (4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6) \\ (5,1) & (5,2) & (5,3) & (5,4) & (5,5) & (5,6) \\ (6.1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6) \end{array}\right]\end{align} So total number of sample points are $$n(S)=6 \times 6=36$$ An odd number as the sum

Let $H=\{an\, odd \,number\, as\, a\, sum \}.$

then from the sample space, we see that $n(H)=18$.

Thus the probability is: $$P(H)=\dfrac{n(H)}{n(S)}=\dfrac{18}{36}=\dfrac{1}{2}$$

## Question 7(ix)

Two dice are thrown simultaneously. Find the probability of getting a multiple of 3 as the sum.

### Solution

The sample space rolling a pair of dice is \begin{align} S&=\{(i, j) ; i, j=1,2,3,4,5,6\}\\ &=\left[\begin{array}{llllll} (1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) \\ (2,1) & (2,2) & (2,3) & (2,4) & (2,5) & (2,6) \\ (3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6) \\ (4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6) \\ (5,1) & (5,2) & (5,3) & (5,4) & (5,5) & (5,6) \\ (6.1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6) \end{array}\right]\end{align} So total number of sample points are $$n(S)=6 \times 6=36$$ A multiple of 3 as the sum

Let $K=\{ a\, multiple\, of\, 3 \,as\, the\, sum \},$

then we see from the sample space that $n(K)=12$.

Thus the probability is: $$P(K)=\dfrac{n(K)}{n(S)}=\dfrac{12}{36}=\dfrac{1}{3}$$

## Question 7(x)

Two dice are thrown simultaneously. Find the probability of getting sum as a prime number.

### Solution

The sample space rolling a pair of dice is \begin{align} S&=\{(i, j) ; i, j=1,2,3,4,5,6\}\\ &=\left[\begin{array}{llllll} (1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) \\ (2,1) & (2,2) & (2,3) & (2,4) & (2,5) & (2,6) \\ (3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6) \\ (4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6) \\ (5,1) & (5,2) & (5,3) & (5,4) & (5,5) & (5,6) \\ (6.1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6) \end{array}\right]\end{align} So total number of sample points are $$n(S)=6 \times 6=36$$ Sum as a prime number

Let $M= \{sum\, as\, a\, prime\, number \}$

then see from the sample space that $n(M)=15$.

Thus the probability of sum as a prime number is: $$P(M)=\dfrac{n(M)}{n(S)}=\dfrac{15}{36}=\dfrac{5}{12}$$

### Go To