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Aptitude Probability Practice Q&A-Easy

2831.A letter is randomly taken from English alphabets. What is the probability that the letter selected is not a vowel?
5/25
2/25
5/26
21/26
Explanation:

Total number of alphabets,$ n\left(S\right)$ = 26

Total number of characters which are not vowels, $n\left(E\right)$ = 21

$\text{P(E) = }\dfrac{\text{n(E)}}{\text{n(S)}} = \dfrac{21}{26}$

2833.In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize?
$ \dfrac{1}{10} $
$ \dfrac{2}{5} $
$ \dfrac{2}{7} $
$ \dfrac{5}{7} $
Explanation:

P [getting a prize] =$ \dfrac{10}{(10 + 25)} $=$ \dfrac{10}{35} $=$ \dfrac{2}{7} $.

2834.A card is drawn from a pack of 52 cards. The probability of getting a queen of club or a king of heart is:
$ \dfrac{1}{13} $
$ \dfrac{2}{13} $
$ \dfrac{1}{26} $
$ \dfrac{1}{52} $
Explanation:

Here, $ n \left(S\right)$ = 52.

Let E = event of getting a queen of club or a king of heart.

Then, $ n \left(E\right)$ = 2.

$\therefore P\left(E\right)$ =$ \dfrac{n(E)}{n(S)} $=$ \dfrac{2}{52} $=$ \dfrac{1}{26} $.

2835.A letter is chosen at random from the word ASSASSINATION. What is the probability that it is a consonant?
4/13
8/13
7/13
6/13
Explanation:

Total Number of letters in the word ASSASSINATION,$ n\left(S\right)$ = 13

Total number of consonants in the word ASSASSINATION = 7

$\text{P[consonant] = }$$\dfrac{7}{13}$

2840.A bag contains 6 black and 8 white balls. One ball is drawn at random. What is the probability that the ball drawn is white?
$ \dfrac{3}{4} $
$ \dfrac{4}{7} $
$ \dfrac{1}{8} $
$ \dfrac{3}{7} $
Explanation:

Let number of balls = (6 + 8) = 14.

Number of white balls = 8.

P [drawing a white ball] =$ \dfrac{8}{14} $=$ \dfrac{4}{7} $.

2842.Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?
$ \dfrac{1}{2} $
$ \dfrac{2}{5} $
$ \dfrac{8}{15} $
$ \dfrac{9}{20} $
Explanation:

Here, S = {1, 2, 3, 4, ...., 19, 20}.

Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.

$\therefore P\left(E\right)$ =$ \dfrac{n(E)}{n(S)} $=$ \dfrac{9}{20} $.

2844.What is the probability of getting a sum 9 from two throws of a dice?
$ \dfrac{1}{6} $
$ \dfrac{1}{8} $
$ \dfrac{1}{9} $
$ \dfrac{1}{12} $
Explanation:

In two throws of a dice, $ n \left(S\right)$ = $(6 \times 6)$ = 36.

Let E = event of getting a sum ={(3, 6), (4, 5), (5, 4), (6, 3)}.

$\therefore P\left(E\right)$ =$ \dfrac{n(E)}{n(S)} $=$ \dfrac{4}{36} $=$ \dfrac{1}{9} $.

2846.Three unbiased coins are tossed. What is the probability of getting at most two heads?
$ \dfrac{3}{4} $
$ \dfrac{1}{4} $
$ \dfrac{3}{8} $
$ \dfrac{7}{8} $
Explanation:

Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}

Let E = event of getting at most two heads.

Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}.

$\therefore P\left(E\right) $ =$ \dfrac{n(E)}{n(S)} $=$ \dfrac{7}{8} $.

2847.A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is:
$ \dfrac{1}{22} $
$ \dfrac{3}{22} $
$ \dfrac{2}{91} $
$ \dfrac{2}{77} $
Explanation:

Let S be the sample space.

Then, $ n \left(S\right)$ = number of ways of drawing 3 balls out of 15

= 15C3

=$ \dfrac{(15 \times 14 \times 13)}{(3 \times 2 \times 1)} $

= 455.

Let E = event of getting all the 3 red balls.

$\therefore n \left(E\right)$ = 5C3 = 5C2 =$ \dfrac{(5 \times 4)}{(2 \times 1)} $= 10.

$\therefore P\left(E\right)$ =$ \dfrac{n(E)}{n(S)} $=$ \dfrac{10}{455} $=$ \dfrac{2}{91} $.

44448.There are 15 balls numbered 1 to 15, in a bag. If a person selects one at random, what is the probability that the number printed on the ball will be a prime number greater than 5?
3/5
1/5
11/5
2/5
Explanation:

The primes between 5 and 15 are: 7,11,13.
So the probability $\dfrac{3}{15}=\dfrac{1}{5}$

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