Probability Test - 6

1 year = 365 days

A leap year has 366 days

A year has 52 weeks. Hence, there will be 52 Sundays for sure.

52 weeks = 52x7 = 364 days

366 – 364 = 2 days

In a leap year, there will be 52 Sundays and 2 days will be left.

These 2 days can be:{Sunday,Monday}{Monday,Tuesday}{Tuesday,Wednesday}{Wednesday,Thursday}{Thursday,Friday}{Friday,Saturday}{Saturday,Sunday}

Of these total 7 outcomes, the favourable outcomes are 2.

Hence, the probability of getting 53 Sundays = 2/7

Total number of marbles =10

Number of green marbles = 4

Probability of getting green marbles =4/10 = 2/5

P(3 or 5) = P(3) + P(5) - P(3 and 5)

Multiples of 3 = {3,6,9,12,15,18}

Therefore, P(3) = 6/20

Multiples of 5 = {5,10,15,20}

Therefore, P(5) = 4/20 

Multiples of both 3 and 5 = {15}

Therefore, P(3 and 5) =1/20

Therefore, P(3 OR 5) = 6/20 + 4/20 - 1/20 = 920

The probability of winning a game = 0.3,

the probability of losing a game is = 1 - 0.3,= 0.7

The total number of outcomes = 36
Favourable outcomes for getting a doublet = 6 { (1,1) (2,2) (3,3) (4,4) (5,5) (6,6) }
Probability of getting a doublet,P(A) = 6/36

Favorable outcomes for getting a total of 4 =3 { (1,3) (2,2) (3,1) }
Probability of getting a total of 4 ,P(B)= 3/36
Favourable outcomes for getting both sum as 4 and also a doublet =1 {2,2}

Probability of getting both sum as 4 and also a doublet,P(A∩B) =1/36

So,probability of getting a doublet or a total of 4 is calculated using addition theorem of probability 

P(A∪ B) = P(A) + P(B) - P(A∩B) = 6/36 + 3/36 - 1/36 = 8/36 = 2/9

Total number of cards, N=100

Prime numbers less than 20: {2,3,5,7,11,13,17,19} = 8 instances

The probability that the number of the card is a prime number less than 20 ,P(p<20) = 8/100= 2/25

Total number of card in a deck = 52

Card removed king, queen and jack of clubs

Therefore, remaining cards = 52 - 3=49

Therefore, number of favorable outcomes = 49

Number of clubs in a deck in a deck of 52 cards = 13

According to the question, the king, queen and jack of clubs are removed from a deck of 52 playing cards In this case, total number of clubs = 13 - 3 = 10

Therefore, the probability of getting ‘a club’,P(C) = Number of favorable outcomes  / Total number of possible outcomes  = 10/49

Two different dice are thrown simultaneously being number 1, 2, 3, 4, 5 and 6 on their faces.

We know that in a single thrown of two different dice, the total number of possible outcomes is (6 × 6) = 36.

Let E = event of getting a multiple of 2 on one die and a multiple of 3 on the other die.

The events of a multiple of 2 on one die and a multiple of 3 on the other die will be E = [(2, 3), (2, 6), (3, 2), (3, 4), (3, 6), (4, 3), (4, 6), (6, 2), (6, 3), (6, 4), (6, 6)] = 11

Therefore, probability of getting ‘a multiple of 2 on one die and a multiple of 3 on the other die’

P(E) = Number of favourable outcomes /  Total number of possible outcome = 11/36