M.C.Q (1 Marks)

MCQ 1011 Mark

A bag contains 5 red balls and n green balls. If the probability of drawing a green ball is three times that of a red ball, then the value of n is

A
18
✓
15
C
10
D
20

Answer

Correct option: B.

(b) Total number of balls is n + 5 out of which one can be drawn in (n + 5) ways.
∴ Total number of elementary events n + 5
Out of 5 red balls, one ball can be chosen in 5 ways and out of in green balls one green ball can be chosen in ways. Therefore,
$p_1$ Probability of drawing a green ball $=\frac{n}{n+5}$ p2 Probability of drawing a red ball $=\frac{5}{n+6}$
$p_1=3 p_2 \Rightarrow \frac{n}{n+5}=3 \times \frac{5}{n+5} \Rightarrow n=15$

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MCQ 1021 Mark

In a single throw of two dice, the probability of getting 6 as a product of two numbers obtained is

A
$\frac{4}{9}$
B
$\frac{2}{9}$
C
$\frac{1}{9}$
D
$\frac{5}{9}$

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MCQ 1031 Mark

Two dice are rolled together. The probability of getting sum of numbers on the two dice is 2, 3 or 5, is

✓
$\frac{7}{36}$
B
$\frac{11}{36}$
C
$\frac{5}{36}$
D
$\frac{4}{9}$

Answer

Correct option: A.

$\frac{7}{36}$

(a) When two dice are rolled together, there are 36 elementary events associated to this random experiment. Sum of the numbers on two dice is 2 or 3 or 5, if we get one of the following outcomes: (1, 1), (1, 2), (2, 1), (1, 4), (4, 1), (2, 3), (3, 2)
∴ Favourable number of elementary events = 7
Hence, required probability = $\frac{7}{36}$

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MCQ 1041 Mark

Aarushi and Avni visit a particular office in the same week (Monday to Friday). Each is equally likely to visit the office on any one day as on another. The probability that both will visit the office on two consecutive days is

A
$\frac{11}{25}$
✓
$\frac{8}{25}$
C
$\frac{7}{25}$
D
$\frac{9}{25}$

Answer

Correct option: B.

$\frac{8}{25}$

(b) Aarushi and Avni each can visit the particular office on any one of the 5 days of the week. So, the total number of ways of visiting the office by Aarushi or Avni or both, is 5 x 5 = 25. ∴ Total number of elementary events = 25. Aarushi and Avni both can visit the office on two consecutive days in the following ways:

Aarushi:M T W Th	Avni: M T W Th
Avni: T W Th F	Aarushi: T W Th F

$\therefore \quad$ Favourable number of elementary events $=8$
Hence, required probability $=\frac{8}{25}$.

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MCQ 1051 Mark

The probability of selecting a consonant from the lecetrs of the word 'TRIANGLE' is

A
$\frac{2}{7}$
B
$\frac{3}{8}$
✓
$\frac{5}{8}$
D
$\frac{1}{8}$

Answer

Correct option: C.

$\frac{5}{8}$

(c) There are 8 letters in the word 'TRIANGLE' out of which three are vowels and the remaining 5 are consonants
∴Total number of elementary events = 8
Favourable number of elementary events = 5
Hence, required probability =$\frac{5}{8}$.

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MCQ 1061 Mark

One ticket is drawn from a bag containing 70 tickets numbered 1 to 70. The probability that the drawn ticket bears a number which is a multiple of 5 or 7, is

A
$\frac{1}{10}$
B
$\frac{1}{70}$
C
$\frac{6}{70}$
✓
$\frac{11}{35}$

Answer

Correct option: D.

$\frac{11}{35}$

(d) Out of 70 tickets, one ticket can be chosen in 70 ways.
∴ Total number of elementary events = 70
There are 14 tickets bearing numbers which are multiple of 5 and 10 tickets bear numbers which are multiple of 7. Out of these two tickets are marked with numbers which are multiple of 5 and 7 both i.e. multiple of 35. So, there are 14+10-2-22 tickets bearing numbers which are either multiple of 5 or multiple of 7.
∴ Favourable number of elementary events = 22.
Hence, required probability $=\frac{22}{70}=\frac{11}{35}$.

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MCQ 1071 Mark

Hundred cards marked with numbers 1 to 100 are placed in a box. If a card is selected randomly from the box, then the probability that the number on the card selected is perfect square is

A
$\frac{1}{100}$
B
$\frac{1}{25}$
✓
$\frac{1}{10}$
D
$\frac{9}{10}$

Answer

Correct option: C.

$\frac{1}{10}$

(c) One card can be selected from 100 cards in 100 ways. So, total number of elementary events = 100.
There are 10 cards marked with numbers 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 which are perfect squares, out of these one card can be selected in 10 ways.
∴ Favourable number of elementary events = 10
Hence, required probability = $\frac{10}{100}=\frac{1}{10}$.

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MCQ 1081 Mark

A three digit number is chosen at random, the probability that its hundred's digit, ten's digit and unit's digit are consecutive integers in descending order, is

A
$\frac{1}{75}$
B
$\frac{4}{225}$
✓
$\frac{2}{225}$
D
$\frac{1}{45}$

Answer

Correct option: C.

$\frac{2}{225}$

(c) There are 900 three digit numbers, namely, 100, 101, ..., 998, 999. Out of these one number can be chosen in 900 ways.
∴ Total number of elementary events = 900.
Three digit numbers having hundred's, ten's and unit's digit as consecutive integers in descending order are 210, 321, 432, 543, 654, 765, 876, 987. These are 8 numbers, out of which one can be chosen in 8 ways.
∴ Favourable number of elementary events = 8
Hence, required probability =$\frac{8}{900}=\frac{2}{225}$.

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MCQ 1091 Mark

x = ABCDEFGHIJ ... Z. A letter is selected at random from the letters in odd positions. The probability that it is a vowel, is

✓
5/13
B
6/13
C
7/13
D
8/13

Answer

Correct option: A.

5/13

(a) There are 13 letters in odd positions, one of these letters can be chosen in 13 ways.
∴ Total number of elementary events = 13 .
Vowels A, E, I, O, U are at 1, 5, 9, 15 and 21 positions respectively. So all vowels are at odd positions.
∴ Favourable number of elementary events = 5.
Hence, required probability = $\frac{5}{13}$.

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MCQ 1101 Mark

A three digit number is chosen at random, the probability that it is divisible by both 2 and 3, is

A
$\frac{1}{8}$
B
$\frac{1}{9}$
✓
$\frac{1}{6}$
D
$\frac{1}{12}$

Answer

Correct option: C.

$\frac{1}{6}$

(c) Three digit numbers are. 100, 101, ..., 999. These are 900 numbers out of which one can be chosen in 900 ways. So, total number of elementary events is 900.
A number will be divisible by both 2 and 3, if it is divisible by 6. Three digit numbers divisible by 6 are 102, 108, 114, ...., 996. These numbers form an AP with first term = 102, common difference d = 6 and last term = 996. Let there number be n. Then, $996=102+(n-1) \times 6 \Rightarrow 894=6(n-1) \Rightarrow n-1=149 \Rightarrow n=150$
So, there are 150 numbers between 102 and 996 which are divisible by 6. Out of these one number can be chosen in 150 ways.
Hence, required probability $=\frac{150}{900}=\frac{1}{6}$.

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MCQ 1111 Mark

A number is chosen at random from the numbers -5,-4,-3,-2,-1,0,1,2,3,4,5. Then, the probability that square of this number is less than or equal to 1, is

A
$\frac{9}{11}$
✓
$\frac{3}{11}$
C
$\frac{8}{11}$
D
$\frac{7}{11}$

Answer

Correct option: B.

$\frac{3}{11}$

(b) There are 11 numbers, out of which a number can be chosen in 11 ways.
So, total number of elimentary events (outcomes) = 11.
The square of the chosen number will be less than or equal to 1, if the chosen number is any one of the following numbers - 1, 0, 1.
∴ Favourable number of elementary events = 3.
Hence, required probability = 3/11

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Maths M.C.Q (1 Marks)