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- Question 11 Review Exercise 6
- determine the probability. ====Solution==== Total number colors $$n(S)=4$$ P(orange) The orange color cove
- Question 9 & 10 Review Exercise 6
- on $=100,0000$. First we are computing the total number of ways arranging these digits using repeated per... {3 ! \cdot 2 !}=420 $$ But we have find the total number that are greater than $1$ million. In this case number should not start with $0$, therefore the total ... ular men sit together. ====Solution==== The total number of ways that $n$ persons can be seated around a c
- Question 5 & 6 Review Exercise 6
- it next to each other? ====Solution==== The total number of seats are six so $$n=6$$ The total different ... ng next to each other? ====Solution==== The total number of seats are six so $n=6$. The total different a... the two seats like one seat, and hence the total number of arrangements round the circle in this case are... ted next to each other ====Solution==== The total number of ways sitting of six people around a circular t
- Question 7 & 8 Review Exercise 6
- d with the digits $0,1,2,3,4,5,6,7,8,9$ if each number starts with $35$ and no-digits appear more than once? ====Solution==== If each telephone number starts with $35.$ It means have to fill the fir
- Question 3 & 4 Review Exercise 6
- . x . y . y . y . z . z . z . z . z $$ The total number of letter in this word are twelve, so $n=12$ out... m_2=3$, and five are $z$, so $m_3=5$. Thus total number of ways that $x^4 y^3 z^5$ can be arrange are \be
- Question 1 Review Exercise 6
- ">(c): $28$ </collapse> iii. How many six digits number can be formed from the digits $\{1,2,3,4,6,7,8\}$... collapsed="true">(d): $4775$</collapse> vii. The number of all possible matrices of order $3 \times 3$ wi
- Question 10 Exercise 6.5
- apples or both are good? ====Solution===== Total number of Apples $=20$ number of Oranges $=10$ number of defective apples $=5$ number of defective oranges $=3$. Totál good apples $=15$ Defective apples $=
- Question 8 Exercise 6.5
- right]\end{align} The probability that the sum of number is $7$ is: $$=\dfrac{6}{36}$$ The probability that the sum of number is $11$ is: $$=\dfrac{2}{36}$$ The events that th
- Question 5 and 6 Exercise 6.5
- t is the probability that the second dice shows a number greater than $4$? ====Solution==== The sample spa... \\ n(S)&=6 \end{align} Let \begin{align}A&=\{ a\, number\, greater\, tha, 4\}\\ &=\{5,6\}\\ \text{then} n(... hus, the probability that the second dice shows a number greater than $4$ is: \begin{align}P(A)&=\dfrac{n(
- Question 3 and 4 Exercise 6.5
- selected at random. Find the probability that its number is either odd or the square of an integer. ====So... &=\{1,3,5,..,29\}\\ n(A)&=15\\ \text{Let}\, B&=\{ number \,square \,of\, an\, integer \}\\ &=\{1,4,9,16,25... \cap B)=\dfrac{3}{30}$$ The probability that the number is either odd or the square of an integer is: \be
- Question 7 Exercise 6.4
- 5) & (6,6) \end{array}\right]\end{align} So total number of sample points are $$n(S)=6 \times 6=36$$ doub... nce the possibility of getting doublet of an even number is: $$P(A)=\dfrac{n(A)}{n(S)}=\dfrac{3}{36}=\dfra... 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
- Question 6 Exercise 6.4
- ards $=52$ Drawing either space or hearts Total number of spade in deck of cards are $13$ total number of hearts are also $13$ Thus the probability of gettin... Total cards $=52$ Drawing a diamond card. Total number of diamond cards in deck of eards is $13.$ Hence... l cards $=52$(iv) Drawing a face card. The total number of face cards in deck of card are $12.$ Thus the
- Question 5 Exercise 6.4
- of $3$ men and $2$ women. ====Solution==== Total number of persons $=6+4=10$. Total number of ways to select $5$ out of these $10$ are: \begin{align}{ }^{10)} C_... $2$ women. By multiplication principle the total number of ways, selecting $3$ men and $2$ worncn are: \b... of $2$ men and $3$ women. ====Solution==== Total number of persons $=6+4=10$. Total number of ways to sel
- Question 2 Exercise 6.4
- bility that all are green? ====Solution==== Total number of balls are: $4+5+6=15$ balls Total number of ways drawing three balls at random are: $${ }^{15} C_3=\d... !}{(15-3) ! 3 !}=455 $$ All are green The total number ways of favorable outcomes for green balls are: $... bility that all are white? ====Solution==== Total number of balls are: $4+5+6=15$ balls Total number of w
- Question 1 Exercise 6.4
- ling a dice. What is the probability of rolling a number less than $1$ ? ====Solution==== Rolling a number less than $1$ Let \begin{align}B&=\{\}\\ &=\phi \text{t... ling a dice. What is the probability of rolling a number greater than $0$ ? ====Solution==== Rolling a number greater than $0$ Let \begin{align}C&=\{1,2,3,4,5,6\}