Question

1. The birthday of six random people has been checked. Find the probability that

(a) At least one of them is born in September.

(b) All five are born in the Spring. Spring here means one of the month March, April, or May.

(c) At least two of them are born in the same month. In this problem you can assume that a year is 365 days.

2.A fair die is rolled three times. We say that a match has occurred if the outcome of the first throw is 1, or the outcome of the second throw is 2, or the outcome of the third throw is 3. Find the probability of the event that a match occurs.

3. An ordinary deck of playing cards is randomly divided into two parts, each containing at least one card.

(a) What is the probability that each part contains at least one ace.

(b) Find the probability that each part contains exactly two aces.

Answer 1

Answer:

1.

a)

Given,

To give the probability that at least one of them is born in September

i.e.,

P(At least one of them born in September) = 1 - P(none born in September)

= 1 - (11^6 / 12^6)

= 1 - (1771561/2985984)

= 1 - (0.5933)

= 0.4067

Required probability = 0.4067

b)

To determine the probability that all six of them were born in September

P(All six of them born in September) = 3^6 / 12^6

= 729 / 2985984

= 0.0002

Required probability = 0.0002

c)

To determine the probability that at least two of them were born in same month)

Required probability = 1 - P(none were born in same month)

= 1 - (12p7 / 12^6)

= 1 - (12*11*10*9*8*7 / 12^6)

= 1 - (665280/2985984)

= 1 - 0.2228

Required probability = 0.7772

3.

Here we have that each card has 2 different ways of dissemination either in gathering 1 or 2. In the event that gatherings are expected unmistakable, at that point there will be 2 situations when every one of the cards goes to aggregate 1 or all gathering 2 which will be prohibited.

So we have total number of ways = 2^52 - 2

a)

To determine the probability that each part contains at least one ace

Here we have 4 ace cards so that we may receive at least one ace = 2^4 - 2

Number of ways of distribution to others = 2^(52-4) = 2^48

Now number of positive ways = 2^48(2^4 - 2)

Now required probability = positive ways/total number of ways

= 2^48(2^4 - 2) / 2^52 - 2

Required probability = 0.875

b)

To determine the probability that each part contains exactly two aces

Here number of ways of distributing the 2 aces to both groups is given as

= 4! / (2!*2!)

= 24/4

= 6

Number of ways of distribution of remaining cards = 2^48

Now required probability = 6*2^48 / 2^52-2

= 0.375

So required probability = 0.375

I hope it helps you

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