**Q1. On the average, how may times must a die be thrown until one gets a 6?**

Solution:

Let p be the probability of a 6 on a given trial. Then the probabilities of success for the first time on each trial are (if q = 1-p):

Trail Probability of success on trial

1 p

2 pq

3 pq^2

The sum of probability is

p+pq+pq^2+.....

= p(1+q+q^2+......)

= p/(1-q)

= 1

The mean number of trials, m, is

m = p+2pq+3pq^2+4pq^3+........ (1)

To apply the usual trick for summing a geometric series:

qm = pq+2pq^2+3pq^3+..... (2)

Subtracting (2) from (1) gives:

m - qm = p+pq+pq^2+.....

= 1

mp = 1

m = 1/p

Since p=1/6 here

**m = 6**

**Q2**

**WHAT IS THE PROBABILITY THAT I FLIP THIS PENNY 5 TIMES, IT WILL COME UP HEADS AT LEAST 2 TIMES?The trick here is the phrase “at least 2.” This means: what is the probability that of 5 flips, I will end up with 2 or 3 or 4 or 5 heads? Think of this as the inverse to the problem: what is the probability that of 5 flips, I see a head only once or no times at all?**

The probability out of 5 flips of getting 0 heads:

= (probability of getting 5 tails)

= (0.5)5

= 3.125%

The probability out of 5 flips of getting 1 head:

=(probability of getting 4 tails) × (probability of getting 1 head) × 5C1

= (0.5)4 × (0.5)1 × 5C1

= (0.5)4 × (0.5)1 × 5! ÷ (1! × 4!)

= 15.625%

The probability of getting 0 or 1 heads in 5 flips = 3.125% + 15.625% = 18.750%.

The probability of getting 2 or more heads in 5 flips = 1 – 18.75% = 81.25%.

**Q3. BIRTHDAY PROBLEM: WHAT’S THE PROBABILITY THAT IN A ROOM FULL OF K PEOPLE, AT LEAST 2 PEOPLE WILL HAVE THE SAME BIRTHDAY?**

Each person has 365 “possible” birthdays (ignoring leap years); in a room of

*k*people, the total number of possible birthdays is 365

*k*.

In order for at least 2 people to have the same birthday, this means that 2 or 3 or 4….have the same birthday, which is the “inverse” or complement of none of the

*k*birthdays in the group being the same. This latter event is a permutation; if John is born on January 1 and Jeff is born on December 31, this is a different outcome than John being born on December 31 and Jeff being born on January 1 (i.e., position matters.) The total number of ways to choose

*k*different birthdays from 365 elements (with no repetitions) is 365! ÷ (365 –

*k*)!. So for a room full of

*k*people, the probability that at least 2 have the same birthday is:

1 – [365! ÷ (365 –

*k*)!] ÷ 365

*k*

When

*k*= 15, the probability is 25.3%; when

*k*= 50, the probability rises to 97.0%!

**Q4 WHAT IS BAYES’ THEOREM AND WHEN IS IT USED?**

Bayes’ theorem gives the relationship between the probabilities of Event A and Event B, and the conditional probability of Event A occurring given the occurrence of Event B, written P (A|B).

For example, assume a drug test is correct 99% of the time, meaning that when someone has used the drug, they test positive for the drug 99% of the time; when someone hasn’t used the drug, they test negative 99% of the time. If 2% of the population uses the drug, what is the probability that someone who has tested positive actually does use the drug?

Prob(drug user) = 2%

Prob(test positive | drug user) = 99%

Prob(test positive | not drug user) = 1%

So we have:

Prob(drug user |test positive) =

[Prob(test positive |drug user) × Prob(drug user)] ÷ {[Prob(test positive |drug user) × Prob(drug user)] + [Prob (test positive |not drug user) × Prob(not drug user)]}

Therefore:

Prob(drug user |test positive) = [99% × 2%] / [(99% × 2%) + (1% × 98%)] = 66.9%

So for a positive test, there’s only a 2/3 chance that the test-taker is a drug user.