Aptitude Probability Practice Q&A-Easy

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

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

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

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} $.

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}$

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

Number of white balls = 8.

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

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} $.

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} $.

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} $.

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} $.

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