Aptitude Tests 4 Me

Basic Numeracy/Quantitative Aptitude

Detailed Solution

177. c: Let S = Sample space
A = the event of drawing ‘2’ white balls.
B = the event of drawing ‘2’ red balls.
A B = The event of drawing 2 white balls or 2 red balls.
i.e. the event of drawing ‘2’ balls of same colour.
n(S) = 10C2 = 10/ (2x 8) = 45
n(A) = 6C2 = 6/ ((2x 4) = (6 x 5 ) / 2 = 15
n(B) = 4C2 = 4/ (2x 2) = (4x3) / 2 = 6
P(A) = n(A) / n(S) = 15/45 = 1/3
P(B) = n(B) / n(S) = 6/45 = 2/15
P(A B) = P(A) + P(B)
= 1/3 + 2/15 = (5+2) / 15
P(A B) = 7/15

178. c: Let ‘E1’, ‘E2’, ‘E3’ be the events of selections of A, B, and C respectively. Let P(E3) = x
P(E2) = 3. P(E3) = 3x
and P(E1) = 2P(E2) = 2 x 3x = 6x
As there are only ‘3’ candidates ‘A’, ‘B’ and ‘C’ we have to select at least one of the candidates A or B or C, surely.
P( E1 E2 E3) = 1
and E1, E2, E3 are mutually exclusive.
P(E1 E2 E3) = P(E1) + P(E2) + P(E3)
1 = 6x + 3x + x
10x – 1 or x = 1/10
P(E3) = 1/10, P(E2) = 3/10 and P(E1) = 6/10 = 3/5

179. d: Total possible orders = starters x mains x deserts =4x5x3=60

180. c: The probability of calling correctly when a coin is tossed is p=1/2. The probability of getting it correctly three times is P=(1/2)x(1/2)x(1/2)=1/8

181. d: The probability of getting the first number correct is 1/5
The probability of getting the second number correct is ¼ since there are only 4 left
The probability of getting the third number correct is 1/3 since there are only 3 left
The probability of getting the fourth number correct is 1/2 since there are only 2 left
The probability of getting the fifth number correct is 1/1 since there are only 1 left
The probability of getting all five numbers is 1/5x1/4x1/3x1/2x1/1=1/120
In other words the number of permutations is simply factorial 5 or 5!

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