Probability Test - 1

Probability that both are red cards, without replacement = (26/52) x (25/51) = 0.240

The probability of choosing the first purse = 1/2
The probability of choosing coin in first purse = 4/7
Now, the probability that the coin is picked from the first purse = (1/2) × (4/7)

The probability of choosing the second purse = 1/2
The probability of choosing coin in second purse = 6/8
Now, the probability that the coin is picked from the first purse = (1/2) × (6/8)
Required probability = (1/2) x (4/7) + (1/2) x (6/8) = 2/7 + 3/8 = 37/56

The answer is 6/7. It does not matter, whether the year is a leap year, or not. But, the day selected can be any day from Sunday to Saturday.

Hence, the probability that the day will not be a Tuesday is 6/7.

The probability of winning a match is 0.2.
If the person wins 1 match and looses 2 matches, then the probability = 0.2 × (0.8)² = 0.128
If the person wins 2 matches and looses 1 match, then the probability = (0.2)² × 0.8 = 0.032
If the person wins 3 matches, then the probability = (0.2)³ = 0.008
Therefore, the probability that he will win at least one of the three matches = 0.128 + 0.032 + 0.008
= 0.168

Probability that both are red cards with numbers less than or equal to 10, without replacement = (18/52) x (17/51) = 0.115

Integers from 1 to 50, which are not multiples of 2 or 3 or 10, are 1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47 and 49.
So, required number of integers = 50 - 17 = 33
Probability = 33/50

Possible outcomes are {(1, H), (2, H), (3, H), (4, H), (5, H), (6, H), (1, T), (2, T), (3, T), (4, T), (5, T), (6, T)}
Favourable outcomes = {(2, H), (4, H), (6, H)}
Hence, required probability = Number of favourable outcomes/Total number of outcomes

Required Probability = 3/12 = 1/4

As five cards – ten, jack, queen, king and ace of diamonds fall down, there are only 47 cards left in the pack, with 9 face cards.

Therefore, P(a face card) = 9/47