# Probability

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• Q1

The probability that in a group of n persons all will have their birthdays on different days of the year is:

Marks:1

##### Explanation:

• Q2

Cards are drawn one by one without replacement from a pack of 52 cards. The probability that 10 cards will precede the first ace is

Marks:1

492/12495

##### Explanation:

Total number of cards = 52
Number of aces = 4
Probability of one ace = 4/52
Probability of othere ten cards taken
one by one without replacement = (48/51)(47/50)(46/49)...(39/42)
Therefore,
Required Probability =
(48/51)(47/50)(46/49)...(39/42).(4/52)
= 492/12495

• Q3

In 50 tickets of a lottery there are three prizes first, second and third. The probability that a person having 5 tickets will win a price is

Marks:1

##### Explanation:

• Q4

The probability of getting 10 in a single throw of three fair dice is

Marks:1

1/8.

##### Explanation:

Exhaustive no. of cases = 63

10 can appear on three dice either as distinct number as following:

(1, 3, 6) ; (1, 4, 5); (2, 3, 5) and  each can occur in 3! ways.

Or 10 can appear on three dice as repeated digits as following :

(2, 2, 6), (2, 4, 4), (3, 3, 4) and each can occur in  3!/ 2! ways.

No. of favourable cases  = 33!+3 (3!/ 2!) =27

Hence, the required probability= 27/ (63) = 1/8.

• Q5

Two dice are thrown simultaneously. The probability of obtaining a total score of seven is

Marks:1

1/6.

##### Explanation:

When two dice are thrown then there are 6 × 6 exhaustive cases  n = 36.

Let A denote the event “total score of 7” when 2 dice are thrown then

A = [(1, 6), (2, 5), (3, 4), (4, 3),(5, 2), (6, 1)].

Thus there are 6 favourable cases.

m = 6

By definition P(A) = m/n

P(A) = 6/36 = 1/6