Question

A fisherman's probability of a successful day of fishing is 0.4. Given that he fished five days every week.

a.) Categorize each random variable as discrete or continuous                              

b.) Define what type of distribution involved                                                        

c.) Find the probability that in any week he has i) exactly & ii) at least three successful days 

Answer 1

a) For a given sample space $S$ of some experiment, a random variable is any rule that associates a number with each outcome in $S.$

Let $X=$ the number of successful day of fishing in a week.

A discrete random variable is a random variable whose possible values either constitute a finite set or else can be listed in an infinite sequence in which there is a first element, a second element, and so on (“countably” infinite).

Hence $X$ is a discrete random variable.

b) $X=$ the number of successful day of fishing in a week.

1. The experiment consists of a sequence of $n$ trials, where $n$ is fixed in advance of the experiment: $n=7.$

2. Each trial can result in one of the same two possible outcomes: success or failure.

3. The trials are independent.

4. The probability of success $P(S)$ is constant from trial to trial: $p=0.4.$

Then $X,$ the number of successes in $n=7$ trials is the binomial random variable:

$X\sim Bin(n, p).$

$$P(X=x)=\dbinom{n}{x}p^x(1-p)^{n-x}, x=0,1,2,...,n$$

c)

Given $n=7, p=0.4, q=1-p=1-0.4=0.6$

i)

$$P(X=3)=\dbinom{7}{3}(0.4)^3(1-0.4)^{7-3}$$

$$=35(0.4)^3(0.6)^{4}=0.290304$$

ii)

$$P(X\geq3)=1-P(X<3)$$

$$=1-P(X=0)-P(X=1)-P(X=2)$$

$$=1-\dbinom{7}{0}(0.4)^0(1-0.4)^{7-0}-\dbinom{7}{1}(0.4)^1(1-0.4)^{7-1}$$

$$-\dbinom{7}{2}(0.4)^2(1-0.4)^{7-2}=1-(0.6)^7$$

$$=1-(0.6)^7-7(0.4)(0.6)^6-21(0.4)^2(0.6)^5$$

$$=0.580096$$