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- Question 3 Exercise 6.4
- ains eight questions. If a student guesses the answer for each question, find the probability that $8$ answers are correct. ====Solution==== We have $8$ questions, each question has two options. Therefore, The... ual outcome to occur is $$\dfrac{1}{256}$$ $8$ answers are correct. Let $$A=\{8\}$$ Obviously only on
- Question 1 Review Exercise 6
- correct option. <panel> i. In how many ways can we name the vertices of pentagon using any five of t... ) $4880$ \\ <btn type="link" collapse="a1">See Answer</btn><collapse id="a1" collapsed="true">(a): $25... ts $\{1,2,3,4,5,6,7\}$ if repeated digits are allowed? * %%(a)%% $14$ * (b) $42$ * %%(c)... (d) $21$ \\ <btn type="link" collapse="a2">See Answer</btn><collapse id="a2" collapsed="true">(c): $28
- Question 7 and 8 Exercise 6.2
- rom the digits $1,2,3,4$ and 5 if repetitions allowed? ====Solution==== There are three places (hundre... lled by five digits, Moreover repetition is allowed. Hence $E_1$ occurs in $m_1=5$ ways only $E_2$... digits $1,2,3,4$ and 5 if repetitions are not allowed? ====Solution==== If repetition is not allowed then each digit can appear once in each number. In t
- 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 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 "No," because there are no three of the given points ... 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 answ
- 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 7 and 8 Exercise 6.3
- d{align} =====Question 8===== A student is to answer $7$ out of $10$ questions in an examination. How many choices has he, if he must answer the first $3$ questions? ====Solution==== Option... choose different $7$ questions out of $10$ to answer are: $${ }^{10} C_7=\dfrac{10 !}{(10-7) ! 7 !}=120$$ If he must answer the first three, then the remaining questions a
- 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 5 and 6 Exercise 6.2
- he digits $2,4,5,7,9$ ? (Repetition not being allowed). 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 with the remaining 4 , third place with the rema
- 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 $n^
- Question 3 & 4 Exercise 6.1
- +\dfrac{3}{8 !}=\dfrac{75}{8 !}$ ====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