Four Probability Questions 1 In the game of bridge, 13 cards are dealt with each

Question

Four Probability Questions

1 In the game of bridge, 13 cards are dealt with each player from a standard deck. A hand which contains no aces and no cards above nine is called a Yarborough. This is named after British nobleman Charles Worseley, the second Earl of Yarborough, who is said to bet a thousand to one against dealing with such a hand. What is the probability of a Yarborough ? What do you think of his bet?

2 Suppose there are N students in a class.

a). What is the probability that at least two students in the class have the same birthday?

b). How large does the class have to be to make this probability at least 1/2?

3 If 4 married couples are arranged in a row, what is the probability that no husband sits next to his wife?

4 A jar contains three +1, three -1 and four 0. Let Q(x) := a0 + a1 x + a2 x*x with a0, a1 and a2 drawn from the jar with replacement. a). What is the probability that Q has only one root? b). What is the probability that Q has two real roots? c). Given that Q has two real roots, what’s the probability that both of them are positive?

Explanation / Answer

1. It is given that 13 cards are dealt with each player from a standard deck. The number of ways in which this can be done is 52C13. Now, a Yarborough happens when the hand contains no aces and no cards above nine(52-16=36). The number of ways in which this can happen is 36C13.

Thus, the probability of a Yarborough is given by 36C13/52C13 =0.003639

The probability is very low and hence it can be said that the chances of winning the bet is very less.

2.(a) It is given that there are 'n' students in a class. The birthday of any student can fall on any day out of the 365 days of the year, so the total number of ways in which the birthdays of 'n' students can happen in a year is 365n.

If we consider the situation where at least two students in the class have the same birthday,

P(Atleast two students in the class have the same birtday)= 1 - P(No two students in the class have the same birthday)

If the birhdays of all n-students fall on different days, then the number of fauurable cases are:

365x(365-1)x(365-2).......x[365-(n-1)]

So, P(No two students in the class have the same birthday) = {365x(365-1)x(365-2).......x[365-(n-1)]} / 365n

So, P(Atleast two students in the class have the same birtday) = 1 - {365x(365-1)x(365-2).......x[365-(n-1)]} / 365n

(b) The value of 'n' that will make the probability that at least two students in the class have the same birthday 1/2 is given by:

1 - {365x(365-1)x(365-2).......x[365-(n-1)]} / 365n =1/2

{365x(365-1)x(365-2).......x[365-(n-1)]} / 365n =1/2

By trial and error it can be seen that a class size of 22 or 23 will lead this probability to be approximately 1/2.

3. There are 4 married couples, that is 8 people. The total number of ways in whicg 8 people can be made to sit in a row is 8!=40320.

Suppose we consider a situation where none of the couples sit together. The number of ways in which this can happen is 13824

Thus, the probability of none of the couples sitting together is 13824/40320 = 0.34286

4. The total number of possible ways in which a0, a1, and a2 can be selected is 3x3x3x3= 27

(a) The equation Q will have only one root when a2=0 and a1 0, Thus, This is possible in 3x2 ways i.e. 6 ways.

Thus, the probability is 6/27 =2/9

(b) The discriminant will be positive only when either a2 or a0 is -1 not both or either one or both of them is 0. This is possible in 21 ways. Thus, the probability of this happenening is 21/27 = 7/9

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