If the letters of the word 'ALGORITHM' are arranged at random in a row what is the probability the letters 'GOR' must remain together as a unit?
Number of letters in the word 'ALGORITHM' =9
If 'GOR' remain together, then considered it as 1 group.
$\therefore$ Number of letters $=6+1=7$
Number of word, if 'GOR' remain together $=7$ !
Total number of words from the letters of the word 'ALGORITHM' $=9$ !
$\therefore \quad$ Required probability $=\frac{7!}{9!}=\frac{1}{72}$
Six new employees, two of whom are married to each other, are to be assigned six desks that are lined up in a row. If the assignment of employees to desks is made randomly, what is the probability that the married couple will have non-adjacent desks?
Let the couple occupied adjacent desks consider those two as 1.
There are $(4+1)$ i.e ., 5 persons to be assigned.
$\therefore \quad$ Number of ways of assigning these five person $=5!\times 2!$
Total number of ways of assigning 6 persons $=6$ !
$\therefore$ Probability that the couple has adjacent desk $=\frac{5!\times 2!}{6!}=\frac{2}{6}=\frac{1}{3}$
Probability that the married couple will have non-adjacent desks $=1-\frac{1}{3}=\frac{2}{3}$
If an integer from 1 through 1000 is chosen at random, then find the probability that the integer is a multiple of 2 or a multiple of 9.
Multiple of 2 from 1 to 1000 are $2,4,6,8, \ldots, 1000$
Let $n$ be the number of terms of above series.
$$\begin{aligned} & \therefore \quad n \text {th term }=1000 \\ & \Rightarrow \quad 2+(n-1) 2=1000 \\ & \Rightarrow \quad 2+2 n-2=1000 \\ & \Rightarrow \quad 2 n=1000 \\ & \therefore \quad n=500 \end{aligned}$$
Since, the number of multiple of 2 are 500 .
So, the multiple of 9 are $9,18,27, \ldots, 999$
Let $m$ be the number of term in above series.
$$\begin{aligned} & \therefore \quad m \text { th term }=999 \\ & \Rightarrow \quad 9+(m-1) 9=999 \\ & \Rightarrow \quad 9+9 m-9=999 \\ & \Rightarrow \quad 9 m=999 \\ & \therefore \quad m=111 \end{aligned}$$
Since, the number of multiple of 9 are 111 .
So, the multiple of 2 and 9 both are $18,36, \ldots$, 990
Let $p$ be the number of terms in above series.
$$\begin{array}{lr} \therefore & p \text { th term }=990 \\ \Rightarrow & 18+(p-1) 18=990 \\ \Rightarrow & 18+18 p-18=990 \\ \Rightarrow & 18 p=990 \\ \therefore & p=\frac{990}{18}=55 \end{array}$$
Since, the number of multiple of 2 and 9 are 55 .
$\therefore$ Number of multiple of 2 or $9=500+111-55=556$
$\therefore$ Required probability $=\frac{n(E)}{n(S)}=\frac{556}{1000}=0.556$
An experiment consists of rolling a die until a 2 appears.
(i) How many elements of the sample space correspond to the event that the 2 appears on the $k$ th roll of the die?
(ii) How many elements of the sample space correspond to the event that the 2 appears not later than the $k$ th roll of the die?
In a through of a die there is 6 sample points.
(i) If 2 appears on the $k$ th roll of the die.
So, first $(k-1)$ roll have 5 outcomes each and $k$ th roll results 2 i.e., 1 outcome.
$\therefore$ Number of element of sample space correspond to the event that 2 appears on the $k$ th roll of the die $=5^{k-1}$
(ii) If we consider that 2 appears not later than $k$ th roll of the die, then it is possible that 2 comes in first throw i.e., 1 outcome.
If 2 does not appear in first throw, then outcomes will be 5 and 2 comes in second throw i.e., 1 outcome, possible outcome $=5 \times 1=5$
$$\begin{aligned} &\text { Similarly, if } 2 \text { does not appear in second throw and appears in third throw. }\\ &\begin{aligned} \therefore \quad \text { Possible outcomes } & =5 \times 5 \times 1 \\ \text { Given, } \quad \text { series } & =1+5+5 \times 5+5 \times 5 \times 5+\ldots+5^{k-1} \\ & =1+5+5^2+5^3+\ldots+5^{k-1} \\ & =\frac{1\left(5^k-1\right)}{5-1}=\frac{5^k-1}{4} \end{aligned} \end{aligned}$$
A die is loaded in such a way that each odd number is twice as likely to occur as each even number. Find $P(G)$, where $G$ is the event that a number greater than 3 occurs on a single roll of the die.
It is given that, $2 \times$ Probability of even number $=$ Probability of odd number
$\Rightarrow \quad P(O)=2 P(E)$
$\Rightarrow \quad P(O): P(E)=2: 1$
$\therefore$ Probability of occurring odd number, $P(O)=\frac{2}{2+1}=\frac{2}{3}$
and probability of occurring 5each number, $$P(E)=\frac{1}{2+1}=\frac{1}{3}$$
Now, $G$ be the even that a number greater than 3 occur in a single roll of die. So, the possible outcomes are 4,5 and 6 out of which two are even and one odd.
$$\begin{aligned} \therefore \quad \text { Required probability } & =P(G)=2 \times P(E) \times P(O) \\ & =2 \times \frac{1}{3} \times \frac{2}{3}=\frac{4}{9} \end{aligned}$$