Probability

The probability of landing on heads each time a fair coin is flipped, is 1/2.

Assuming that the question was supposed to be:

"What is the probability of landing on heads twice in a row?"

To calculate compound probabilities like this, we first have to work out the probability of landing on heads each time, and then multiply these two probabilities to get a compound probability.

1/2 x 1/2 = 1/4

So the probability of landing on heads twice in a row = 1/4 (for a fair coin)

P = 0.072314699... ≈ 7.23%

SOLUTION

Let's identify the 8 people by integers 1, 2,..., 8, and their birthday by x1, x2,..., x8.

To include the possibility of xi = Feb. 29 (of a leap year), we will consider 4(365) +

1 = 1451 days, so the probability of a person having Feb. 29 as his birthday is

P(xi = Feb. 29) = 1/1461. The probability of a person having his birthday on any

given day except Feb. 29 is P(xi ≠ Feb. 29) = 4/1461.

Let's first calculate the probability of person 1 and 2 having the same birthday, x1 = x2, and persons 3, 4,..., 8, all have different birthdays x3,...,x8.

Now we go.

We take person 1 and have two possibilities:

A).- x1 = Feb. 29 B).- x1 ≠ Feb. 29

P(x1 = Feb. 29) = 1/1461; P(x ≠ Feb. 29) = 1 - 1/1461

We will construct two branches. One for each possibility.

Branch A: given x1 = Feb. 29, the probability of person 2 having the same birthday

is P(x2 = x1) = 1 - 1/1461. The probability of person 3 not having the same birthday is P(x3 ≠ x1 = x2) = 1 - 1/1461. The probability of person 4 not having either two birthdays is P(x4 ≠ x1, x4 ≠ x3) = 1 - 5/1461. The probability of person 5 not sharing x1 nor x3 nor x4 is P(x5 ≠ x4, x5 ≠ x3, x5 ≠ x2) = 1 - 9/1461. And so a series of terms develops:

P(xi ≠ ... ) = 1 - (4i - 3)/1461 from i = 1, to i = n-2

where n is the number of people in the room (In this particular question for 8

people, n = 8).

The branch we have constructed for possibility A is the product of the individual

conditional probabilities;

A: (1/1461)2 π1n-2[1-(4i-3)/1461]

where π1n-2 is the product operator "pi" of "i" terms from i =1, to i = n-2.

Branch B: Given x1 ≠Feb. 29, the probability of person 2 having the same birthday

is P(x2 = x2) = 4/1461. The probability of person 3 not having the same birthday

is P(x3 ≠ x1) = 1 - 4/1461. The probability of person 4 not sharing birthdays with 2 and 3 is P(x4 ≠ x3, x4 ≠ x2) = 1 - 8/1451. The probability of person 5 not

sharing x2 nor x3 nor x4 is P(x5 ≠ ...) = 1 - 12/1461. And so a series of terms

develops:

P(xi ≠ ... ) = 1- (4i -4)/1461 from i = 2, to i = n-1

where n is the number of people in the room (In this particular question for 8 people, n = 8).

Branch for possibility B: (1-1/1461)(4/1461) π2n-1[1-(4i-4)/1461]

The sum of these two branch expressions give the probability of persons 1 and 2

sharing a birthday in a group of n persons:

P(x1=x2)=(1/1451)2π1n-2[1-(4i-3)/1461]+(1-1/1461)(4/1461)π2n-1[1-(4i-4)/1461]

Now, the probability of any two persons, and only those two persons, to share a birthday in a group of n persons, is the previous probability, times the number of different combinations of two persons, (i,j), in a group of n persons, nC2

(where, nC2 = n!/[2!(n-2)!] )

The final expression is:

--------------------------------------------------------------------------------------------

P = nC2{(1/1451)2π1n-2[1-(4i-3)/1461]+(1-1/1461)(4/1461)π2n-1[1-(4i-4)/1461]}

--------------------------------------------------------------------------------------------

For the case of n = 8, the result is, P = 0.072314699... ≈ 7.23%

The probability of rolling a 4 is 1/6. (There are 6 numbers one of which is "4")

The probability of rolling an odd number is 3/6 = 1/2 (There are 6 numbers 3 of which are odd)

Thus, in one roll of the die, the odds of rolling a 4 oran odd number = 1/6 + 3/6 = 4/6 = 2/3.