Search
You can find the results of your search below.
Fulltext results:
- Question 3 Exercise 6.4
- that $8$ answers are correct. ====Solution==== We have $8$ questions, each question has two options. Th... that $7$ answers are correct. ====Solution==== We have $8$ questions, each question has two options. Th... that $6$ answers are correct. ====Solution==== We have $8$ questions, each question has two options. Th... east $6$ answers are correct. ====Solution==== We have $8$ questions, each question has two options. Th
- Question 5 and 6 Exercise 6.3
- } C_2=66$ lines. The question now is whether we have counted any line twice. the the answer is "No," ... oints 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
- Question 5 and 6 Exercise 6.2
- . How many of these are even? ====Solution==== We have to fill four places with these five digits $2,4,5... rs Out of these for even number, the unit digit have to be filled by $2$ or $4$. So, we are left with
- 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 $n=9$
- Question 1 Exercise 6.3
- {align} But $n$ can not be negative therefore, we have $n=9$. =====Question 1(ii)===== Solve $^{n+1} C_... {align} But $n$ can not be negative, therefore we have $n=8$. =====Question 1(iii)===== Solve $n^2 C_2
- Question 7 Exercise 6.5
- ints therefore, by addition law of probability we have \begin{align} P(A \cup B)&=P(A)+P(B) \\ & =\dfrac... ther red nor king so by complementary events, we have that probability that it is neither red nor king
- Question 7 & 8 Review Exercise 6
- ach telephone number starts with $35.$ It means have to fill the first two places with these two digits, then the remaining digits are: $$10-2=8$$ We have to fill the remaining four places with these $8$
- Question 9 Exercise 6.2
- umber of them used at a time? ====Solution==== We have to form signals using the six colors of the six f
- Question 8 Exercise 6.5
- ive, therefore by addition law of probability we have \begin{align}P(A \cup B)&=P(A)+P(B)\\ &=\dfrac{
- Question 9 Exercise 6.5
- {4}{5}\end{align} None will be selected. Here we have to find probability that none of both is selected
- Question 10 Exercise 6.5
- $=22$ Now two fruits are chosen at random and we have to find the probability that either both are appl
- Question 9 & 10 Review Exercise 6
- n as: $$=\dfrac{7 !}{3 ! \cdot 2 !}=420 $$ But we have find the total number that are greater than $1$ m
- Question 11 Review Exercise 6
- green event. By addition law of probability, we have \begin{align}\boldsymbol{P}( Red or Green )&=P(\t