Probability (Lecture-02)
Important Formulae and Shortcuts
1. When ’ $n$ ’ fair dice are rolled, the number of favourable cases to get sum $r=$ coefficient of $x^{r}$ in $\left(\mathrm{x}^{1}+\mathrm{x}^{2}+\ldots . . \mathrm{x}^{6}\right)^{\mathrm{n}}($ multinomial theorem $)$
and $P($ sum $=k)=$ coefficient $x^{r}$ in $\dfrac{\left(x^{1}+x^{2}+\ldots . . x^{6}\right)^{n}}{6^{n}}$
Note:-
i. In case of 2 fair dice, number of favourable cases to get sum $r=\left\{\begin{array}{l}r-1 ; \text { for } 2 \leq \mathrm{r} \leq 7 \\ 13-r ; \text { for } 8 \leq \mathrm{r} \leq 12\end{array}\right.$
ii. When 3 fair dice are rolled, number of favourable cases to get sum $r$
$= \begin{cases}\dfrac{1}{2}(\mathrm{r}-1)(\mathrm{r}-2) & ; \text { for } 3 \leq \mathrm{r} \leq 18 \\ 25 & ; \text { for } \mathrm{r}=9,12 \\ 27 & ; \text { for } \mathrm{r}=10,11 \\ \dfrac{1}{2}(20-\mathrm{r})(19-\mathrm{r}) ; \text { for } 13 \leq \mathrm{r} \leq 18\end{cases}$
2. When ’ $n$ ’ fair coins are tossed, the probability of getting
i. exactly r heads (tails) $=\dfrac{{ }^{n} C _{\mathrm{r}}}{2^{n}}$
ii. atleast one head (tails) $=1-\dfrac{1}{2^{\mathrm{n}}}$
3. If $r$ squares are selected at random from a chess board. The probability that they lie on a
diagonal = $\left\{\begin{array}{cl} \dfrac{2}{{ }^{64} \mathrm{C} _8} & \text {;if } \mathrm{r}=8 \\ \dfrac{4\left\{{ }^7 \mathrm{C} _{\mathrm{r}}+\ldots+{ }^{\mathrm{r}} \mathrm{C} _{\mathrm{r}}\right\}+2\left({ }^8 \mathrm{C} _{\mathrm{r}}\right)}{{ }^{64} \mathrm{C} _{\mathrm{r}}} & \text {; if } 1 \leq \mathrm{r} \leq 7\end{array}\right.$
4. Let $n(A)=n$. Then
i. $P(r$ subsets of A selected at random have no common element $)=\left(\dfrac{2^{r}-1}{2^{r}}\right)^{n}$
ii. P(2 subsets of A selected at random have r elements in common $)={ }^{\mathrm{n}} \mathrm{C} _{\mathrm{r}}\left(\dfrac{3}{4}\right)^{\mathrm{n}-\mathrm{r}}\left(\dfrac{1}{4^{\mathrm{r}}}\right)$
iii. $\mathrm{P}(\mathrm{C} \cup \mathrm{B}=\mathrm{A})=\left(\dfrac{3}{4}\right)^{\mathrm{n}}$ if $\mathrm{B} \& \mathrm{C}$ are two subsets of $\mathrm{A}$.
5. Out of $n$ pairs of socks, if $\mathrm{r}(\mathrm{r}<\mathrm{n})$ socks are selected at random, the probability that there is no pair $=\dfrac{{ }^{n} C _{r} 2^{r}}{{ }^{2} C _{r}}$
6. If $\mathrm{n}$ letters are put in $\mathrm{n}$ addressed envelopes at random, then the probability that
i. Exactly one letter may be in wrong envelope is 0
ii. At least one letter may be in wrong envelope $=1-\mathrm{P}($ all are in right envelope $)=1-\dfrac{1}{\mathrm{n} !}$
iii. Exactly $r(\neq 1)$ letters one is wrong envelope $={ }^{n} P _{r}\left(1-\dfrac{1}{1 !}+\dfrac{1}{2 !}-\dfrac{1}{3 !}+\ldots \ldots .+\dfrac{(-1)^{r}}{r !}\right)$
7. If $\mathrm{A} _{1}, \mathrm{~A} _{2} \ldots . . \mathrm{A} _{\mathrm{n}}$ play a game in which $\mathrm{P}$ (success $)=\mathrm{p}$ and $\mathrm{P}($ failure $)=\mathrm{q}$, then
i. $P\left(A _{1}\right.$ is winning $)=\dfrac{p}{1-q^{n}}$
ii. $P\left(A _{2}\right.$ is winning $)=\dfrac{q p}{1-q^{n}}$
iii. $P\left(A _{3}\right.$ is winning $)=\dfrac{q^{2} p}{1-q^{n}}$
iv. $P\left(A _{4}\right.$ is winning $)=\dfrac{q^{n-1} p}{1-q^{n}}$ where $1 \leq \mathrm{k} \leq \mathrm{n}$
8. Suppose a word is given. Let the number of vowels be $\mathrm{V}$, number of consonants be $\mathrm{C}$ and the total number of letters be T. If the letters of the word are arranged at random, then the probability that.
i. relative positions of vowels $\&$ consonants do not change $=\dfrac{\mathrm{V} ! \mathrm{C} \text { ! }}{\mathrm{T} !}$
ii. The order of the vowels do not change $=\dfrac{1}{\text { number of ways of arranging the vowels }}$
iii. The order of the consonants do not change $=\overline{\text { number of ways of arranging consonants }}$
iv. The order of the vowels as well as consonants do not change
$=\dfrac{1}{(\text { number of ways of arranging the vowels)(number of ways of arranging consonants) }}$
Solved Examples
1. If two events $\mathrm{A}$ and $\mathrm{B}$ are such that $\mathrm{P}\left(\mathrm{A}^{1}\right)=0.3, \mathrm{P}(\mathrm{B})=0.4$ and $\mathrm{P}\left(\mathrm{A} \cap \mathrm{B}^{1}\right)=0.5$, then $\mathrm{P}\left(\dfrac{\mathrm{B}}{\mathrm{A} \cup \mathrm{B}^{1}}\right)=$
(a) 0.9
(b) 0.5
(c) 0.6
(d) 0.25
Show Answer
Solution: $\mathrm{P}\left(\dfrac{\mathrm{B}}{\mathrm{A} \cup \mathrm{B}^{\prime}}\right)=\dfrac{\mathrm{P}\left(\mathrm{B} \cap\left(\mathrm{A} \cup \mathrm{B}^{\prime}\right)\right)}{\mathrm{P}\left(\mathrm{A} \cup \mathrm{B}^{\prime}\right)}=\dfrac{P(B \cap A)}{P\left(A \cup B^{\prime}\right)}$
$=\dfrac{\mathrm{P}(\mathrm{A})-\mathrm{P}\left(\mathrm{A} \cap \mathrm{B}^{\prime}\right)}{\mathrm{P}(\mathrm{A})+\mathrm{P}\left(\mathrm{B}^{\prime}\right)-\mathrm{P}\left(\mathrm{A} \cap \mathrm{B}^{\prime}\right)} \quad=\dfrac{0.7-0.5}{0.7+0.6-0.5}=\dfrac{1}{4}$
Answer: d
2. If points $x$ and $y$ are selected at random from the intervals $[0,2]$ and $[0,1]$ the probability that $y \leq x^{2}$ is
(a) $\dfrac{1}{2}$
(b) $\dfrac{2}{3}$
(c) $\dfrac{3}{4}$
(d) $\dfrac{1}{4}$
Show Answer
Solution:
Required probability $=\dfrac{\text { Area of shaded region }}{\text { Area of rectangel } \mathrm{OABC}}$
$=\dfrac{\int _{0}^{1} x^{2} \mathrm{dx}+1 \mathrm{x} 1}{2 \mathrm{x} 1}=\dfrac{\dfrac{4}{3}}{2}=\dfrac{2}{3}$
Answer: b
3. If the integers $m$ and $n$ are selected between 1 and 100 , the probability that a number of the form $7^{\mathrm{m}}+7^{\mathrm{n}}$ is divisible by 5 is
(a) $\dfrac{1}{4}$
(b) $\dfrac{1}{7}$
(c) $\dfrac{1}{8}$
(d) $\dfrac{1}{49}$
Show Answer
Solution:
Form of m | Digit at unit’s place | Digit at units place | Form of n |
---|---|---|---|
$7^{4 \lambda \lambda}$ | 1 | 9 | $7^{4 \lambda+2}$ |
$7^{4 \lambda+1}$ | 7 | 3 | $7^{4 \lambda+3}$ |
$7^{4 \lambda+2}$ | 9 | 1 | $7^{4 \lambda}$ |
$7^{4 \lambda+3}$ | 3 | 7 | $7^{4 \lambda+1}$ |
Clearly for 25 values of $m$, there are 25 values of $n$ between $0 \& 100$
$\therefore \quad$ Favourable ways is $4(25 \times 25)=2500$
$\therefore \quad$ required probability is $\dfrac{2500}{100 \times 100}$ (as total number of ways is $100 \times 100$ )
$=\dfrac{1}{4}$
Answer: a
4. Two persons each makes a single throw with a pair of dice. The probability that their throws are unequal is
(a) $\dfrac{1}{216}$
(b) $\dfrac{70}{216}$
(c) $\dfrac{115}{216}$
(d) none of these
Show Answer
Solution: Required probability
$=\quad 1-\mathrm{P}($ their throws are equal $)$
$=1-\left\{\dfrac{1^{2}}{36^{2}}+\dfrac{2^{2}}{36^{2}}+\dfrac{3^{2}}{36^{2}}+\ldots . .+\dfrac{6^{2}}{36^{2}}+\dfrac{5^{2}}{36^{2}}+\ldots . .+\dfrac{1^{2}}{36^{2}}\right\}$
$=1-\dfrac{146}{36^{2}}$
$=1-\dfrac{73}{648}$
$=\dfrac{575}{648}$
Answer: d
5. A coin is tossed $n$ times. If the probability of getting atleast on head is greater that of getting atleast two tails by $\dfrac{5}{32}$, then $\mathrm{n}$ is equal to
(a) 5
(b) 10
(c) 15
(d) none of these
Show Answer
Solution: $\mathrm{P}(\mathrm{X} \geq 1)-\mathrm{P}(\mathrm{Y} \geq 2)=\dfrac{5}{32}$
Let $\mathrm{X}$ be a rendom variable denoting number of heads and $\mathrm{Y}$ denote the number of tails.
$\begin{aligned} & =\left(1-{ }^{\mathrm{n}} C _{0}\left(\dfrac{1}{2}\right)^{0}\left(\dfrac{1}{2}\right)^{\mathrm{n}-0}\right)-\left(1-{ }^{\mathrm{n}} \mathrm{C} _{0}\left(\dfrac{1}{2}\right)^{0}\left(\dfrac{1}{2}\right)^{\mathrm{n}}-{ }^{\mathrm{n}} \mathrm{C} _{1}\left(\dfrac{1}{2}\right)^{1}\left(\dfrac{1}{2}\right)^{\mathrm{n}-1}\right)=\dfrac{5}{32} \\ & \Rightarrow\dfrac{n}{2^n} =\dfrac{5}{32} \text {, comparing LHS and RHS we get }\\ & \Rightarrow \mathrm{n}=5 \end{aligned}$
Answer: a
6. Five digit numbers are to bc formed using the digits $0,2,4,6,8$ without repetition. The probability that the number is divisible by 20 is
(a) $\dfrac{1}{2}$
(b) $\dfrac{1}{3}$
(c) $\dfrac{1}{4}$
(d) $\dfrac{2}{5}$
Show Answer
Solution: The last two places should be $20,40,60,80$
$\therefore \quad$ Favourable cases $=4.3$ !
Hence required probability
$=\dfrac{4.3 !}{4.4 !}=\dfrac{1}{4}$
Answer: c
Exercise
1. A natural number $x$ is chosen at random from the first one hundred natural numbers. The probability $\dfrac{(x-20)(x-40)}{(x-30)}<0$ that is
(a) $\dfrac{1}{50}$
(b) $\dfrac{3}{50}$
(c) $\dfrac{7}{25}$
(d) $\dfrac{9}{50}$
Show Answer
Answer: c2. A wire of length $\ell$ is cut into three pieces. What is the probability that the three pieces form a triangle is
(a) $\dfrac{1}{2}$
(b) $\dfrac{1}{4}$
(c) $\dfrac{2}{3}$
(d) none of these
Show Answer
Answer: b3. If $X \& Y$ are independent binomial variates $B\left(5, \dfrac{1}{2}\right)$ and $\left(7, \dfrac{1}{2}\right) B$ then $P(X+Y=3)$ is
(a) $\dfrac{55}{1024}$
(b) $\dfrac{55}{4098}$
(c) $\dfrac{55}{2048}$
(d) none of these
Show Answer
Answer: a4. Two persons A \& B agree to meet at a place between 11 to 12 noon. The first one to arrive waits for 20 minutes and then leave. If the time of their arrival be independent and at random, then the probability that $\mathrm{A} \& \mathrm{~B}$ meet is
(a) $\dfrac{1}{3}$
(b) $\dfrac{2}{3}$
(c) $\dfrac{4}{9}$
(d) $\dfrac{5}{9}$
Show Answer
Answer: d5. The altitude through $\mathrm{A}$ of $\triangle \mathrm{ABC}$ meets $\mathrm{BC}$ at $\mathrm{D}$ and the circumscribed circle at $\mathrm{E}$. If $\mathrm{D}$ is $(2,3)$, $E(5,5)$, the ordinate of the orthocentre lies on the lines is $y=1, y=2, y=3$ $\mathrm{y}=10$ is
(a) $\dfrac{2}{5}$
(b) $\dfrac{3}{5}$
(c) $\dfrac{4}{5}$
(d) $\dfrac{1}{7}$
Show Answer
Answer: b6. A sum of money is rounded off to the nearest rupee, the probability that the rounded off error is atleast ten paise is
(a) $\dfrac{1}{100}$
(b) $\dfrac{9}{100}$
(c) $\dfrac{49}{100}$
(d) $\dfrac{81}{100}$
Show Answer
Answer: d7. The decimal parts of two numbers taken at random are found to be six places. Probability that second can be subtracted from first one without borrowing is
(a) $\left(\dfrac{1}{2}\right)^{6}$
(b) $\left(\dfrac{9}{20}\right)^{6}$
(c) $\left(\dfrac{11}{20}\right)^{6}$
(d) $\left(\dfrac{3}{20}\right)^{6}$
Show Answer
Answer: c8. Two small squares on a chess board are choose at random. Probability that they have a common side is
(a) $\dfrac{1}{3}$
(b) $\dfrac{1}{9}$
(c) $\dfrac{1}{18}$
(d) $\dfrac{5}{18}$
Show Answer
Answer: c9. Match the following:
Column I | Column II | ||
---|---|---|---|
(a) | If 4 fair dice are rolled, the probability of getting the sum 10 , is | (p) | $7 / 15$ |
(b) | If $x \& y$ are selected from the set $\{1,2,3 \ldots 10\}$, the probability that $\mathrm{x}^{2}-\mathrm{y}^{2}$ is divisible by 3 is | (q) | $2 / 7$ |
(c) | If 10 men are sitting in a row of seats, the probability of choosing 3 of them so that no two of them are from adjacent seats is | (r) | $8 / 15$ |
(d) | Triangles are formed with vertices of a regular octagon. If a triangle is chosen at random, the probability that it does not have any side common with the octagon, is | (s) | $5 / 81$ |
Show Answer
Answer: $\mathrm{a} \rightarrow \mathrm{s} ; \mathrm{b} \rightarrow \mathrm{r} ; \mathrm{c} \rightarrow \mathrm{p} ; \mathrm{d} \rightarrow \mathrm{q}$10. A triangle is formed with the vertices of $n$ sided regular polygon. Then the probability that the triangle.
On the basis of above information, answer the following questions
i. Have exactly one side common with the side of the polygon is
(a) $\dfrac{6(\mathrm{n}-2)(\mathrm{n}-3)}{\mathrm{n}(\mathrm{n}-1)}$
(b) $\dfrac{6(n-3)}{(n-1)(n-2)}$
(c) $\dfrac{6\quad (n-4)}{(n-1)(n-2)}$
(d) $\dfrac{6(n-2)}{(n-1)(n-3)}$
ii. Have exactly two sides common with the side of the polygon is
(a) $\dfrac{3}{(n-1)(n-2)}$
(b) $\dfrac{6(n-4)}{(n-1)(n-2)}$
(c) $\dfrac{6(n-3)}{(n-1)(n-2)}$
(d) $\dfrac{6}{(n-1)(n-2)}$
iii. Does not have any side common with the side of the polygon is
(a) $\dfrac{(n-2)(n-3)}{(n-1)(n-2)}$
(b) $\dfrac{(\mathrm{n}-4)(\mathrm{n}-5)}{(\mathrm{n}-1)(\mathrm{n}-2)}$
(c) $\dfrac{(n-1)(n-3)}{n(n-2)}$
(d) $\dfrac{(\mathrm{n}-2)(\mathrm{n}-4)}{\mathrm{n}(\mathrm{n}-3)}$
iv. Have at least one side common with the sides of the polygon is
(a) $\dfrac{6(n-3)}{(n-1)(n-2)}$
(b) $\dfrac{6(n-4)}{(n-1)(n-2)}$
(c) $\dfrac{6(\mathrm{n}-2)}{\mathrm{n}(\mathrm{n}-1)}$
(d) $\dfrac{6(\mathrm{n}-3)}{\mathrm{n}(\mathrm{n}-1)}$
v. Have at most one side common with the sides of the polygon is
(a) $\dfrac{(\mathrm{n}-4)(\mathrm{n}+2)}{(\mathrm{n}-1)(\mathrm{n}-2)}$
(b) $\dfrac{(n-4)(n+1)}{(n-1)(n-2)}$
(c) $\dfrac{(n-3)(n+1)}{(n-1)(n-2)}$
(d) $\dfrac{(n-3)(n-4)}{(n-1)(n-2)}$