**Def. of Probability:**Probability is the measure of uncertainty of any event (any phenomenon happened or bound to happen)**Experiment:**Any phenomenon like rolling a dice, tossing a coin, drawing a card from a well-shuffled deck, etc.**Outcome:**The Result of any event; like number appearing on a dice, side of a coin, drawn out card, etc.**Sample Space:**The set of all possible outcomes.**Event:**Any combination of possible outcomes or the subset of sample space; like getting an even number on rolled dice, getting a head/tail on a flipped coin, drawing out a king/queen/ace of any suit.**Probability = (Number of a Favourable outcome) / (Total number of outcomes)**

or,**P = n (E) / n (S)**

where,

P(A) is the probability of an event “A”

n(E) is the number of favourable outcomes

n(S) is the total number of events in the sample space**Odds in Favour of the Event:**Odds in the favor of any event is the ratio of the number of ways that an outcome can occur to the number of ways it cannot occur.**Odds Against the Event:**Odds against any event is the ratio of the number of ways that an outcome cannot occur to the number of ways it can occur.**Probability Range:**0 ≤ P(A) ≤ 1**Rule of Addition:**P(A∪B) = P(A) + P(B) – P(A∩B)**Rule of Complementary Events:**P(A’) + P(A) = 1**Disjoint Events:**P(A∩B) = 0**Independent Events:**P(A∩B) = P(A) . P(B)**Conditional Probability:**P(A | B) = P(A∩B) / P(B)**Bayes Formula:**P(A | B) = P(B | A) . P(A) / P(B)

A. 20

B. 36

C. 55

D. 60

Now, in half of these ways M can finish before N

A. 1/455

B. 2/455

C. 4/455

D. 1/91

Probability that all the three marbles picked at random are blue = 3C

A. 1/5

B. 1/7

C. 3/7

D. 3/8

A. 1/5

B. 1/6

C. 1/7

D. 1/8

A. 3/7

B. 4/7

C. 5/7

D. 6/7

A. 1/3

B. 1/5

C. 1/6

D. 1/7

A. 12/35

B. 15/35

C. 18/35

D. 21/35

Probability = 3/5 * 4/7 + 2/5 * 3/7

= 12/35 + 6/35 = 18/35

A. 19/52

B. 17/52

C. 4/13

D. 5/13

P(SuK) = 13/52 + 4/52 – 1/52 = 4/13

A. 1/64

B. 1/36

C. 5/9

D. 5/12

Further the three dice can fall in 6 * 6 * 6 = 216 ways

Hence the probability is 6/216 = 1/36

A. 17/91

B. 33/91

C. 51/91

D. 65/91

When 4 marbles are picked at random, then the probability that none is blue is = 12C

=>(12 * 11 * 10 * 9)/(15 * 14 * 13 * 12) = 33/91

A. 13/21

B. 10/19

C. 11/15

D. Cannot be determined

A. 1/10

B. 1/5

C. 3/10

D. 9/10

Consider the 2 books taken as a pair then, number of favourable ways of getting these two books together is 9! 2!

Required probability = 1/5

A. 0.4869

B. 0.5834

C. 0.5904

D. 0.6234

= 1 – P('Losing on all tickets')

= 1 – (0.8)^4 = (1 + (0.8)^2)(1 – (0.8)^2)

= (1.64)(0.36) = 0.5904

A. 1/2

B. 2/5

C. 8/15

D. 9/20

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

P(E) = n(E)/n(S) = 9/20

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