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#### A school has five houses A, B, C, D and E. A class has 23 students, 4 from house A, 8 from house B, 5 from house C, 2 from house D and rest from house E. Asingle student is selected at random to be the class monitor. The probability that theselected student is not from A, B and C is (A) $\frac{4}{23}$                 (B) $\frac{6}{23}$                   (C) $\frac{8}{23}$                   (D) $\frac{17}{23}$

Solution.  Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a
random event. The value is expressed from zero to one.
Total students = 23
Students in A, B, C = 4 + 8 + 5 = 17
Students in C, D = 23 – 17 = 6
Number of favourable cases = 6
Let A be the event that the student is not from A, B, C

$p\left ( A \right )= \frac{Student\, from\, C,D}{Total\, students}$

$p\left ( A \right )= \frac{6}{23}$

#### Someone is asked to take a number from 1 to 100. The probability that it is a prime is (A) $\frac{1}{5}$                     (B) $\frac{6}{25}$                   (C) $\frac{1}{4}$                     (D) $\frac{13}{50}$

Solution.      Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total number of cases = 100
prime number from 1 to 100 = 2, 3, 5, 7, 9, 11, 13, 17, 19,
23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 83, 89, 97
Total prime numbers from 1 to 100 = 25
Probability of getting prime number = $\frac{prime\, no.\, from\, 1\, to\, 100}{Total\, number}$

$\\=\frac{25}{100}\\\\=\frac{1}{4}$

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#### One ticket is drawn at random from a bag containing tickets numbered 1 to 40. The probability that the selected ticket has a number which is a multiple of 5 is(A) $\frac{1}{5}$                     (B) $\frac{3}{5}$                     (C) $\frac{4}{5}$                     (D) $\frac{1}{3}$

Solution.  Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total tickets = 40
Number of tickets multiple of 5 = 5, 10, 15, 20, 25, 30, 35, 40
Total favourable cases = 8
Let A be the event of getting a ticket with number multiple of 5.
p(A) = $\frac{Number\, of\, tickets\, the\, bought}{Total\, tickets}$
$p\left ( A \right )= \frac{8}{40}= \frac{1}{5}$

#### A girl calculates that the probability of her winning the first prize in a lottery is 0.08. If 6000 tickets are sold, how many tickets has she bought? (A) 40              (B) 240            (C) 480            (D) 750

Solution.     Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total cases = 6000
Probability of getting first prize (p(A)) = 0.08
p(A) $= \frac{Number\, of\, tickets\, the\, bought}{Total\, tickets}$
0.08 × 6000 = Number of tickets the bought

$\frac{8}{100}\times 6000$ = Number of tickets the bought
Number of tickets the bought = 480.

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#### The probability of getting a bad egg in a lot of 400 is 0.035. The number of bad eggs in the lot is (A) 7                (B) 14              (C) 21              (D) 28

Solution.    Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Let event A is to get a bad egg.
So, p (A) = 0.035        (given)
P(A) = $\frac{Number\, of\, favorable\ cases }{Total\, number\, of\, cases}$
0.035 = $\frac{Number\, of\, favorable\ cases }{400}$
Number of favourable cases = $\frac{35}{1000}\times 400= \frac{140}{10}= 14$

#### A card is drawn from a deck of 52 cards. The event E is that the card is not an ace of hearts. The number of outcomes favourable to E is (A) 4                (B) 13              (C) 48              (D) 51

Solution.  Total number of cards = 52
Ace of hearts = 1
The card is not an ace of hearts = 52 – 1 = 51
The number of outcomes favourable to E = 51

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#### When a die is thrown, the probability of getting an odd number less than 3 is (A)   $\frac{1}{6}$                  (B)  $\frac{1}{3}$                   (C)  $\frac{1}{2}$                   (D) 0

Solution.    Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total no. of cases = 6
odd number less than 3 = 1
Number of favorable cases = 1
Probability = $= \frac{1}{6}$

#### The probability that a non leap year selected at random will contain 53 Sundays is (A) $\frac{1}{7}$                    (B) $\frac{2}{7}$                     (C) $\frac{3}{7}$                     (D) $\frac{5}{7}$

Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
In 365 days there are 52 weeks and 1 day.
If it contain 53 sunday then the 1 day of the year must be sunday.
But there are total 7 days.
Hence total number of favorable cases = 1
Hence probability of 53 sunday = $\frac{Number\, of\, favourable\, cases}{Total\, cases}$

$= \frac{1}{7}$

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#### A card is selected from a deck of 52 cards. The probability of its being a red face card is (A) $\frac{3}{26}$             (B) $\frac{3}{13}$             (C) $\frac{2}{13}$              (D) $\frac{1}{2}$

Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total number of cases = 52
Red face cards = 6
Favorable cases = 6
Let event A is to select a card from 52 card.
Probability that it is a red card is p(A)
$P\left ( A \right )= \frac{Number\, of\, favourable\, cases}{Total\, number\, of\, cases}$
$P\left ( A \right )=\frac{6}{52}= \frac{3}{26}$

#### If P(A) denotes the probability of an event A, then (A) P(A) < 0         (B) P(A) > 1       (C) 0 ≤ P(A) ≤ 1          (D) –1 ≤ P(A) ≤ 1

Solution.      Probability; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
(A) P(A) < 0
It is not represent the probability of event A because probability of an event can never be less than 0.
(B) P(A) > 1
It is not represent the probability of event A because probability of an event can never be greater than 1.
(C) 0 ≤ P(A) ≤ 1
It represent probability of event A because probability of an event is always lies from 0 to 1.
(D) –1 ≤ P(A) ≤ 1
It is not represent the probability of event A because probability of an event can never be equal to -1.

Hence option (C) is correct .