3 Marks Question

Question 13 Marks

A bag contains 8 red and 5 white balls. Three balls are drawn at random. Find the Probability that:

All the three balls are white.
All the three balls are red.
One ball is red and two balls are white.

Answer

Three balls are draw at random
$\therefore\ \text{n(S)}={^{13}\text{C}_3}$

P(All the three balls are white)

$=\frac{{^5\text{C}_{13}}}{{^{13}{\text{C}}_3}}=\frac{\frac{5!}{3!2!}}{\frac{13!}{3!10!}}=\frac{10}{\frac{13\times12\times11}{6}}=\frac{5}{143}$

P(All the three balls are red)

$=\frac{{^8\text{C}_{3}}}{{^{13}{\text{C}}_3}}=\frac{\frac{8\times7\times6}{6}}{\frac{13\times12\times11}{6}}=\frac{8\times7\times6}{13\times12\times11}=\frac{28}{143}$

P(One ball is red and two balls are white)

$=\frac{{^8\text{C}_{1}}\times^5\text{C}_2}{{^{13}{\text{C}}_3}}=\frac{8\times10}{\frac{13\times12\times11}{6}}=\frac{40}{143}$

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Question 23 Marks

If the letters of the word ASSASSINATION are arranged at random. Find the Probability that:
Two I’s and two N’s come together.

Answer

We have word 'ASSASSINATION'.
Numberof letter = 13
Letters are 3A's, 4S's, 2I's 1T's and 1O's
Total number of ways these letters can be arranged $=\text{n(S)}=\frac{13!}{3!4!2!2!}$
If 2 I's and 2N's come together, then these as 10 alphabets.
I.e., (IINN), A, A, A, S, S, S, S, T, O
Number of words when 2 I's and 2 N's are come together
$=\frac{10!}{3!4!}\times\frac{4!}{2!2!}$
$\therefore\ \text{Required probability}=\frac{\frac{10!4!}{3!4!2!2!}}{\frac{13!}{3!4!2!2!}}$
$=\frac{4!}{13\times12\times11}=\frac{2}{143}$

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Question 33 Marks

If the letters of the word ASSASSINATION are arranged at random. Find the Probability that:
All A’s are not coming together.

Answer

We have word 'ASSASSINATION'.
Numberof letter = 13
Letters are 3A's, 4S's, 2I's 1T's and 1O's
Total number of ways these letters can be arranged $=\text{n(S)}=\frac{13!}{3!4!2!2!}$
If all A's are coming together, then are 11 alphabets
I.e., (AAA), S, S, S, S, I, I, N, N, T, O
$\therefore$ Number of word when all A's come together $=\frac{11!}{4!2!2!}$
$\therefore$ Probability when all A's come together
$=\frac{\frac{11!}{4!2!2!}}{\frac{13!}{4!3!2!2!}}=\frac{3!}{13\times12}=\frac{1}{26}$
Then the probability thatb all A's does not come together
$=1-\frac{1}{26}=\frac{25}{26}$

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Question 43 Marks

Suppose an integer from 1 through 1000 is chosen at random, find the probability that the integer is a multiple of 2 or a multiple of 9.

Answer

We have integers 1, 2, 3, …., 1000
We have integers 1, 2, 3, ….., 1000
n(S) = 1000
Number of integers which are multiple of 2 = 500 Let the number of integers which are multiple of 9 be n.
n^th term = 999
⇒ 9 + (n -1)9 = 999
⇒ 9 + 9n - 9 = 999
⇒ n = 111
From 1 to 1000, the number of multiples of 9 is 111.
The multiple of 2 and 9 both are 18, 36, …., 990.
Let m be the number of terms in above series.
m^th term = 990
⇒ 18 + (m - 1)18 = 990
⇒ 18 + 18m - 18 = 990
⇒ m = 55
Number of multiples of 2 or 9 = 500 + 111 - 55 = 556 = n(E)
$\therefore\ \text{Required probability}=\frac{\text{n}(\text{E})}{\text{n}(\text{S})}$
$=\frac{556}{1000}=0.556$

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Question 53 Marks

If the letters of the word ASSASSINATION are arranged at random. Find the Probability that:
Four S’s come consecutively in the word.

Answer

We have word 'ASSASSINATION'.
Numberof letter = 13
Letters are 3A's, 4S's, 2I's 1T's and 1O's
Total number of ways these letters can be arranged $=\text{n(S)}=\frac{13!}{3!4!2!2!}$
If for S's come consecutively in the word then we consider these 4S's as 1 group.
So, now number of letters is 10 i.e., (SSSS), A, A, A, I, I, N, N, T,O
$\therefore\ \text{n(E)}=\frac{10!}{3!2!2!}$
$\therefore\ \text{Required probability}=\frac{\frac{10!}{3!2!2!}}{\frac{13!}{3!4!2!2!}}$
$=\frac{4!}{13\times12\times11}=\frac{2}{143}$

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Question 63 Marks

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 nonadjacent desks?
[Hint: First find the probability that the couple has adjacent desks, and then subtract it from 1]

Answer

Number of desk occupied by one couple = 1
Only (4 + 1) = 5 persons to be assigned.
$\therefore$ Number of ways of assigning these 5 persons
5! × 2!
Total number of ways of assigning these 6 persons = 6!
$\therefore$ Probability that a couple has adjacent desk
$=\frac{5!\times2!}{6!}=\frac{1}{3}$
So, the probaility that the married couple will have no-adjacent desks $=1-\frac{1}{3}=\frac{2}{3}$
Hence, the required probability $=\frac{2}{3}$

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Question 73 Marks

An experiment consists of rolling a die until a 2 appears.
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?
[Hint: 1 + 5 + 52 + ... + 5^k-1]

Answer

Number of outcomes when die is thrown is ‘6’
we consider that 2 appears not later than K th roll of the die, then 2 comes before A^th roll.
If 2 appears in first roll, number of ways = 1 If 2 appears in second roll, number of ways = 5 × 1 (as first roll does not result in 2)
If 2 appears in third roll, number of ways = 5 × 5 × 1 (as first two rolls do not result in 2)
Similarly if 2 appears in (k - l)^th roll, number of ways = [5 × 5 × 5 ...… (k - 1) times] × 1 = 5^k-1Possible outcomes if 2 appears before k^th roll = 1 + 5 + 5² + 5³+ ...… + 5^k-l
^{$=\frac{1(5^{\text{k}}-1)}{5-1}=\frac{5^{\text{k}}-1}{4}$}

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Question 83 Marks

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?

Answer

Word ALGORITHM has 9 letters.
If GOR remain together, then it will remain together.
$\therefore$ Number of letter ALGORITHM = 6 + 1 = 7
Number of words = 7!
and the total number of word from ALGORITHM = 9!
So, th required probability $=\frac{7!}{9!}=\frac{7!}{9\cdot8\cdot7!}=\frac{1}{72}$
Hence, the required probability $=\frac{1}{72}$

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Question 93 Marks

One urn contains two black balls (labelled B₁ and B₂) and one white ball. A second urn contains one black ball and two white balls (labelled W₁ and W₂). Suppose the following experiment is performed. One of the two urns is chosen at random. Next a ball is randomly chosen from the urn. Then a second ball is chosen at random from the same urn without replacing the first ball:

Write the sample space showing all possible outcomes.
What is the probability that two black balls are chosen?
What is the probability that two balls of opposite colour are chosen?

Answer

It is given that one of the two urn is chosen, then a ball is randomly chosen from the urn, then a second ball is chosen at random from the same urn without replacing the first ball.

Sample space S = {B₁B₂, B₂B₁, B₁W, WB₁, B₂W, WB₂,BW₁, W₁B, BW₂, W₂B, W₁W₂, W₂W₁}

$\therefore\text{ n(S)}=12$

If two black ball are chosen, then favourable cases are B1B2 and B2B1.

$\therefore\ \text{Reduired probability}=\frac{2}{12}={\frac{1}{6}}$

If two balls of opposite colours are chosen, the favourable cases are B₁W, WB₁, B₂W, WB₂, BW₁, W₁B, BW₂ and W₂B.

$\therefore\ \text{Reduired probability}=\frac{8}{12}={\frac{2}{3}}$

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Question 103 Marks

If the letters of the word ASSASSINATION are arranged at random. Find the Probability that:
No two A’s are coming together.

Answer

We have word 'ASSASSINATION'.
Numberof letter = 13
Letters are 3A's, 4S's, 2I's 1T's and 1O's
Total number of ways these letters can be arranged $=\text{n(S)}=\frac{13!}{3!4!2!2!}$
If no two A's are together, then first we arrange the alphabets other than A's
I.e., S, S, S, S, I, I, N, N, T, O
These letters can be arrangerd in $\frac{10!}{4!2!2!}$ ways.
xS xS xS xS xI xI xN xN xT xOx
Arrangement of these letters creates eleven gaps shown as 'x'.
Three gaps for three A's can be sslected in ¹¹C₃ ways.
$\therefore$ Total number of words when no two A's together
$={^{11}\text{C}_3}\times\frac{10!}{4!2!2!}=\frac{11!}{3!8!}\times\frac{10!}{4!2!2!}$
$\therefore$ The probability that no two A's come together
$=\frac{\frac{11!\times10!}{3!8!4!2!2!}}{\frac{13!}{4!3!2!2!}}=\frac{10!}{8!\times13\times12}=\frac{10\times9}{13\times12}=\frac{15}{26}$

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MATHS 3 Marks Question

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