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- Question 1 and 2 Exercise 6.5
- ac{1}{2}$.\\ Find $P(A \cap B)$. ====Solution==== We know by addition law of probability \begin{align}... ign} Substituting $P(A), P(B)$ and $P(A \cup B)$, we get $$P(A \cap B)=\dfrac{2}{5}+\dfrac{2}{5}-\dfra... dfrac{3}{4}$. Find $P(A \cap B)$ ====Solution==== We are given: $$P(A)=\dfrac{1}{2}, P(\bar{B})=\dfrac{5}{8}, P(A \cup B)=\dfrac{3}{4}$$ We know by complementary events $$P(B)=1-P(\bar{B})$
- Question 7 Exercise 6.4
- number less than 6$\}$, then from sample space, we see that $n(B)=10$. Thus the probability of gett... $ a sum mure than 7$\}$, then from sample space, we see that $n(C)=5$. Thus the probability of getti... sum greater than 10$\}$, then from sample space, we get $n(D)=3$. Thus the probability of getting nu... {a$ sum at least 10$\}$, then from sample space, we see that $n(E)=6$. Thus the probability of getti
- Question 5 and 6 Exercise 6.3
- {12} C_2=66$ lines. The question now is whether we have counted any line twice. the the answer is "... l points are eight so $n=12$. Now for triangles, we have ${ }^{12} C_3=220$ ways to choose the vertices. Again the question is whether we have counted any triangle twice. Again, the answer is "No." If there were a fourth point in one of these triangles, it
- Question 4 Exercise 6.4
- ty of obtaining all heads? ====Solution==== First we construct a tree diagram to find out the sample s... ty of obtaining two heads? ====Solution==== First we construct a tree diagram to find out the sample s... ity of obtaining one hcad? ====Solution==== First we construct a tree diagram to find out the sample s... taining at least one hcad? ====Solution==== First we construct a tree diagram to find out the sample s
- Question 1 Exercise 6.3
- ==== Solve $^n C_2=36$ for $n$. ====Solution==== We are given: \begin{align}&^n C_2=36\\ & \Rightarro... end{align} But $n$ can not be negative therefore, we have $n=9$. =====Question 1(ii)===== Solve $^{n+1} C_4=6,^{n-1} C_2$ for $n$. ====Solution==== We are given: \begin{align} & { }^{n+1} C_4=6 .^{n-1... end{align} But $n$ can not be negative, therefore we have $n=8$. =====Question 1(iii)===== Solve $n^
- Question 3 Exercise 6.4
- ty that $8$ answers are correct. ====Solution==== We have $8$ questions, each question has two options... y one outcome corresponds to this event, because we can select all question to be correct by one way ... ty that $7$ answers are correct. ====Solution==== We have $8$ questions, each question has two options... ty that $6$ answers are correct. ====Solution==== We have $8$ questions, each question has two options
- Question 3 & 4 Exercise 6.1
- +\dfrac{3}{8 !}=\dfrac{75}{8 !}$ ====Solution==== We are taking the L.H.S of the above given equation.... ac{(n+5) !}{(n+3) !}=n^2+9 n+20$ ====Solution==== We are taking the L.H.S of the above given equation.... -5) !}=\dfrac{12(n !)}{(n-4) !}$ ====Solution==== We are given: \begin{align}\dfrac{n(n !)}{(n-5) !}&=... : \dfrac{(n-1) !}{(n-4) !}=9: 1$ ====Solution==== We are given: \begin{align} \dfrac{n !}{(n-4) !}: \d
- Question 13 Exercise 6.2
- &=37,800 \end{align} Begin with $\mathrm{E}$ If we have to pick the combination of words that begin with $E$. It means we have lixed the first one, and the remaining are ... th $C$ If begin with $E$ and end with $C$, means we fixed the two letters and two place. So, the rem... $2 L^{\prime} s$ are to be kept together, then we shall deal these two letters as single, the rem
- Question 6 Exercise 6.4
- n 6(i)===== If one card is drawn at random from a well shuffled pack of $52$ cards. Then find the prob... 6(ii)===== If one card is drawn at random from a well shuffled pack of $52$ cards. Then find the prob... 6(iii)===== If one card is drawn at random from a well shuffled pack of $52$ cards. Then find the prob... 6(iv)===== If one card is drawn at random from a well shuffled pack of $52$ cards. Then find the prob
- Question 9 & 10 Review Exercise 6
- with the digits $2,3,0,3,4,2,3$? ====Solution==== We know that $1$ million $=100,0000$. First we are computing the total number of ways arranging these d... tion as: $$=\dfrac{7 !}{3 ! \cdot 2 !}=420 $$ But we have find the total number that are greater than ... able are: $(n-1)$ ! If two persons sit together, we shall deal these two as one man, then in this cas
- Question 1 and 2 Exercise 6.2
- lve $^n P_5=56(^n P_3)$ for $n.$ ====Solution==== We are given: \begin{align}^n P_5&=56(^n P_3) \\ \Ri... $^n P_5=9(^{n-1} P_4)$ for $n.$ ====Solution==== We are given: \begin{align} ^n P_5&=9(^{n-1} P_4) \\... ==== Solve $n^2 P_2=600$ for $n$ ====Solution==== We are given: \begin{align}n^2 P_2&=60 \\ \Rightarro
- Question 3 and 4 Exercise 6.2
- nting $^n P_r=n(^{n-1} P_{r-1})$ ====Solution==== We are given that: $$^n P_r=n({ }^{n-1} P_{r-1})$$ We are taking the right hand side of the equation \be... _r=^{n-1} P_r+r(^{n-1} P_{r-1})$ ====Solution==== We are given: $$^n P_r=^{n-1} P_r+r({ }^{n-1} P_{r-1
- Question 5 and 6 Exercise 6.2
- ed). How many of these are even? ====Solution==== We have to fill four places with these five digits $... 2,4,5,7,9$, so that repetition is not allowed. We can fill the first place with 5 digits, second wi... e unit digit have to be filled by $2$ or $4$. So, we are left with $3$ numbers. Thus Unit digit: $E_1
- Question 7 & 8 Review Exercise 6
- (B / A)=0.4$, find $P(A \cap B)$ ====Solution==== We know that: \begin{align} P(B \mid A)&=\dfrac{P(A ... $P(B / A)=0.4$, find $P(A / B)$ ====Solution==== We know that $$P(A \mid B)=\dfrac{P(A \cap B)}{P(B)}... digits, then the remaining digits are: $$10-2=8$$ We have to fill the remaining four places with these
- Question 5 Exercise 6.1
- {n !}=2^n(1.3 .5 \ldots(2 n-1))$ ====Solution==== We are taking L.H.S of the above equation \begin{ali... 3 \cdot 5 \ldots(2 n-1)(2 n+1))$ ====Solution==== We are taking L.H.S of the above equation \begin{ali