Quantitative Ability Tech Test Probability Test 2

Total number of balls = (8 + 7 + 6) = 21.

Let E= event that the ball drawn is neither red nor green

= event that the ball drawn is blue.

$\therefore n \left(E\right)$ = 7.

$\therefore P\left(E\right)$ =$ \dfrac{n(E)}{n(S)} $=$ \dfrac{7}{21} $=$ \dfrac{1}{3} $.

Total number of cards = 52

Total Number of King Cards = 4

$\text{P[King] = }\dfrac{4}{52}$

Total Number of Diamond Cards = 13

$\text{P[Diamond] = }\dfrac{13}{52}$

Total Number of Cards which are both King and Diamond = 1

$\text{P[King and Diamond] = }\dfrac{1}{52}$

Here a card can be both a Diamond card and a King. Hence these are not mutually exclusive events.

[Reference : mutually exclusive events]. By Addition Theorem of Probability, we have

P[King or a Diamond] = P[King] + P[Diamond] – P[King and Diamond]

$= \dfrac{4}{52} + \dfrac{13}{52} – \dfrac{1}{52} = \dfrac{16}{52} = \dfrac{4}{13}$

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

Total number of Vowels in the word ASSASSINATION, $n\left(E\right)$ = 6 ( 3 A, 2 I, 1 O)

Probability for getting a vowel, $P\left(E\right)$ =$\dfrac{\text{n(E)}}{\text{n(S)}} = \dfrac{6}{13}$

Clearly, there are 52 cards, out of which there are 12 face cards.

$\therefore$ P [getting a face card] =$ \dfrac{12}{52} $=$ \dfrac{3}{13} $.

Let S be the sample space.

Then, $ n \left(S\right)$ = ⁵²C₂ =$ \dfrac{(52 \times 51)}{(2 \times 1)} $= 1326.

Let E = event of getting 1 spade and 1 heart.

$\therefore n \left(E\right)$= number of ways of choosing 1 spade out of 13 and 1 heart out of 13

=(¹³C₁ ¹³C₁)

= $\left(13 \times 13\right)$

= 169.

$\therefore P\left(E\right)$ =$ \dfrac{n(E)}{n(S)} $=$ \dfrac{169}{1326} $=$ \dfrac{13}{102} $.

Let the event of the occurrence of a number that is odd be ‘A’ and the event of the occurrence of a number that is less than 5 be ‘B’. We need to find P(A or B).
P(A) = 3/6 (odd numbers = 1,3 and 5)
P(B) = 4/6 (numbers less than 5 = 1,2,3 and 4)
P(A and B) = 2/6 (numbers that are both odd and less than 5 = 1 and 3)
Now, P(A or B)= P(A) + P(B) – P(A or B)
= 3/6 + 4/6 – 2/6
P(A or B) = 5/6.

Probability of choosing 1 chocobar = 4/8 = 1/2
After taking out 1 chocobar, the total number is 7.
Probability of choosing 2nd chocobar = 3/7
Probability of choosing 1 icecream out of a total of 6 = 4/6 = 2/3
So the final probability of choosing 2 chocobars and 1 icecream = 1/2 * 3/7 * 2/3 = 1/7

Let the event of getting a greater number on the first die be G.
There are 5 ways to get a sum of 8 when two dice are rolled = {(2,6),(3,5),(4,4), (5,3),(6,2)}.
And there are two ways where the number on the first die is greater than the one on the second given that the sum should equal 8, G = {(5,3), (6,2)}.
Therefore, P(Sum equals 8) = 5/36 and P(G) = 2/36.
Now, P(G|sum equals 8)= P(G and sum equals 8)/P(sum equals 8)
= (2/36)/(5/36)
= 2/5

Taking the individual probabilities of each number, getting a 2 is 1/6 and so is getting a 5.
Applying the formula of compound probability,
Probability of getting a 2 or a 5,
P(2 or 5) = P(2) + P(5) – P(2 and 5)
==> 1/6 + 1/6 – 0
==> 2/6 = 1/3.

We need to find out P(B or 6)
Probability of selecting a black card = 26/52
Probability of selecting a 6 = 4/52
Probability of selecting both a black card and a 6 = 2/52
P(B or 6)= P(B) + P(6) – P(B and 6)
= 26/52 + 4/52 – 2/52
= 28/52
= 7/13.