====== 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==== [[math-11-kpk:sol:unit06:ex6-4-p6 |< Question 6 ]]