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.

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

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: $$P(A)=\dfrac{n(A)}{n(S)}=\dfrac{3}{36}=\dfrac{1}{12}$$

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

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: $$P(B)=\dfrac{n(B)}{n(S)}=\dfrac{10}{36}=\dfrac{5}{18}$$

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

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: $$P(C)=\dfrac{n(C)}{n(S)}=\dfrac{15}{36}=\dfrac{5}{12}$$

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

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: $$P(D)=\dfrac{n(D)}{n(S)}=\dfrac{3}{36}=\dfrac{1}{12}$$

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

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: $$P(E)=\dfrac{n(E)}{n(S)}=\dfrac{6}{36}=\dfrac{1}{6}$$

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

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: $$P(F)=\dfrac{n(F)}{n(S)}=\dfrac{4}{36}=\dfrac{1}{9}$$

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

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

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

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

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

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

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

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: $$P(M)=\dfrac{n(M)}{n(S)}=\dfrac{15}{36}=\dfrac{5}{12}$$

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