Question

Sample Spaces

1. Suppose S is a uniform sample space with N elements. If E is any possible come and ω is the probability function for S evaluate ω(e).

2. Define a probability function on the set A = {1, 2, 3} such that A is not a uniform sample space.

3. Given the sample space B = {a, b, c} and probability function ω on B. If ω(a) = 0.3, ω({a, b}) = 0.8 then find ω(b) and ω(c).

4. Suppose that only 30% of a birds hatchlings survive their first year. If a bird hatches 7 chicks, what is the probability exactly 3 will survive their first year? What is the probability at most 3 will survive?

Answer 1

1. A sample space in which each of the outcomes has the same chance of occurring is called a UNIFORM SAMPLE SPACE.

Note that each outcome can occur only once and hence they are mutually exclusive and mutually exhaustive. Classical definition of probability holds in a uniform sample space because the outcomes here are mutually exclusive, mutually exhaustive, countable and equally likely.

Here S is a uniform sample space with N elements. E is any outcome . Then by classical definition of probability,

2. Given the set A as

A={1,2,3}

A sample space in which each of the outcomes has the same chance of occurring is called a uniform sample space. So if the probability measures are assigned such that any element in the set is given more weightage than the other elements, then it would not be a uniform sample space.

Consider the probability measure on the set A as

It is hereby visible that none of the events are equally likely. Thus A is no more a uniform sample space.

3. Given the sample space B as B={a,b,c} and is the probability measure on B such that

and

Now being a probability measure on B,

({a,b})+(c)=1

i.e. 0.8+(c)=1

i.e. (c)=0.2

Again being a probability measure on B,

(a)+(b)+(c)=1

i.e. 0.3+(b)+0.2=1

i.e. (b)=0.5

Hence (b)=0.5 and (c)=0.2.

4. Let X be the number of bird's hatchlings which survives in the first year. X follows binomial distribution with parameters 7 and 0.3 as the survival of the hatchlings is indicated as success and the death is indicated as failures. Moreover the survival of the hatchlings are independent of one another. Now the pmf of X is given by

where x ranges discretely from 0 to 7.

Now to calculate the probability that exactly 3 will survive their first year, we take x=3.

Again to calculate the probability that atmost 3 will survive their first year, we take x=0,1,2,3 and then add them.

Probability atmost 3 hatchlings will survive their first year

= P(X=0)+P(X=1)+P(X=2)+P(X=3)

=0.0823543+0.2470629+0.3176523+0.2268945

=0.873964

Hope this will help you. In case of any query, do comment. Thanks.