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- Question 9 Exercise 6.3
- 5,6$ or 7 men. If committee contains two men and then it will obviously contain $6$ women. so in this ... \\ &=21\end{align} If committee contains $3$ men then it will contain $5$ women, in this case the tota... \\\ &=210\end{align} If committee contain $4$ men then it will contain $4$ women, in this case total nu... \\ &=525\end{align} If committee contains $5$ men then it will contain $3$ women, in this case total num
- Question 7 Exercise 6.4
- s than $6$ Let $B=\{$ a number less than 6$\}$, then from sample space, we see that $n(B)=10$. Thus t... more than $7$ Let $C=\{$ a sum mure than 7$\}$, then from sample space, we see that $n(C)=5$. Thus th... than $10$ Let $D=\{$ a sum greater than 10$\}$, then from sample space, we get $n(D)=3$. Thus the pro... at least $10$ Let $E=\{a$ sum at least 10$\}$, then from sample space, we see that $n(E)=6$. Thus th
- Question 9 Exercise 6.2
- e six flags. If each signal consist of one color then total number of signals $=^6 P_1=6$. If each signal consist of two color then total number of signal $s=^6 P_2=30$. If each signal consist of three color then total number of signals $=^6 P_3=120$. If each signal consist of four color then total number of signals $=^6 P_4=360$. If each s
- Question 4 Exercise 6.4
- n}S&=(HHII,HHT.HTH.HTT.THII.THT.TTH,TT7),\\ \text{then} n(S)&=2^3=8\end{align} When all heads. Let $$A=\{H H H\}$$ then $$n(A)=1$$ Hence the probability of obtaining all... }S&=(HHII,HHT.HTH.HTT.THII.THT.TTH,TT7),\\ \text{then} n(S)&=2^3=8\end{align} When two heads Let $$B=\{H T$ T.THT.TTH $\}$$ then $$n(B)=3$$ Hence the probability of gelling two h
- Question 1 Exercise 6.4
- than $1$ Let \begin{align}B&=\{\}\\ &=\phi \text{then}\\ P(B)&=\dfrac{n(B)}{n(S)}\\ &=\dfrac{0}{6}\\ &=... han $0$ Let \begin{align}C&=\{1,2,3,4,5,6\},\text{then}\\ P(C)&=\dfrac{n(C)}{n(S)}\\ &=\dfrac{6}{6}\\ &=... multiple of $3$ Let \begin{align}D&=\{3,6\}\text{then}\\ P(D)&=\dfrac{n(D)}{n(S)}\\ &=\dfrac{2}{6}\\ &=... equal to $4$ Let \begin{align} E &=\{4,5,6\}\text{then}\\ P(E)&=\dfrac{n(E)}{n(S)}\\ &=\dfrac{3}{6}\\ &=
- Question 6 Exercise 6.4
- t random from a well shuffled pack of $52$ cards. Then find the probability of each of the Drawing an ac... t random from a well shuffled pack of $52$ cards. Then find the probability of each of the Drawing eithe... t random from a well shuffled pack of $52$ cards. Then find the probability of each of the Drawing a dia... t random from a well shuffled pack of $52$ cards. Then find the probability of each of the Drawing a fac
- Question 10 Exercise 6.2
- ain two students insist to sit next to each other then these two students will be handled as a single st... in two students refuse to sit next to each other, then the total number ways sitting these students in a... }\\ &=\dfrac{7.6 .5 .4 .3 !}{3 !}\\ &=840\\ \text{then}\quad &6720-840=5880\end{align} ====Go To=
- Question 2 Exercise 6.3
- 2-3 n)(n^2-3 n+2)=840 \end{align} Let $y=n^2-3 n$ then the above last equation becomes \begin{align} & y... er} y&=28, \text{or} y=-30\end{align} When $y=28$ then \begin{align} & n^2-3 n=28 \\ & \Rightarrow n^2-3... , so $n=7$. \begin{align}\text{When} &y=-30 \text{then}\\ & n^2-3 n=-30 \\ & \Rightarrow n^2-3 n+30=0\en
- Question 7 and 8 Exercise 6.2
- ed? ====Solution==== If repetition is not allowed then each digit can appear once in each number. In th... vowels. If all the vowels are to kept together, then we shall deal all the vowels as a single alphabet
- Question 1 Review Exercise 6
- 0$</collapse> ix. If $P(A)=\dfrac{1}{2}, P(B)=0$ then $P(A \mid B)$ is: * (a) $0$ * (b) $\dfr... and $B$ are events such that $P(A / B)=P(B / A)$ then * (a) $A \subset B$ but $A \neq B$ * (b
- Question 5 & 6 Review Exercise 6
- If two students refuse to sit next to each other, then the total possible arrangements are: $120-24=96$ ... 20$$ Is Faisal Saima will sit next to each other, then we shall treat both of them as single. In this c
- Question 13 Exercise 6.2
- f two $2 L^{\prime} s$ are to be kept together, then we shall deal these two letters as single, the
- Question 7 and 8 Exercise 6.3
- ! 7 !}=120$$ If he must answer the first three, then the remaining questions are $7$ out of which he h
- Question 3 Exercise 6.4
- }{256}$$ $7$ answers are correct Let $$B=\{7\}$$ then possible outcome or to select $7$ answers correct
- Question 5 and 6 Exercise 6.5
- , number\, greater\, tha, 4\}\\ &=\{5,6\}\\ \text{then} n(A)&=2\end{align} Thus, the probability that th