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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}

Answer.     [B]
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}

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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}

Answer.      [C]
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}

Answer.   [A]
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}

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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

Answer.    [C]
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

Answer.        [B]
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

 

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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

Answer.          [D]
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

Answer.     [A]
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} 
 

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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}

Answer.    [A]
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}

Answer.      [A]
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}

 

 

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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

Answer.          [C]
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 .

 

 

 

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