Let X denote the number of Canon digital cameras sold during...

Let $X$ denote the number of Canon digital cameras sold during a particular week by a certain store. The pmf of $X$ is $\begin{array}{c|ccccc}{x} & {0} & {1} & {2} & {3} & {4} \\ \hline p_{X}(x) & {.1} & {.2} & {.3} & {.25} & {.15} \\ \hline\end{array}$ Sixty percent of all customers who purchase these cameras also buy an extended warranty. tell $Y$ denote the number of purchasers during this week who buy an extended warranty. (a) What is $P(X=4, Y=2) ?[$Hint: This probability equals $P(Y=2 | X=4) \cdot P(X=4) ;$ now think of the four purchases as four trials of a binomial experiment, with success on a trial corresponding to buying an extended warranty 1 (b) Calculate $P(X=Y)$ (c) Determine the joint pmf of $X$ and $Y$ and then the marginal pmf of $Y .$

Let $X$ denote the number of Canon digital cameras sold during a particular week by a certain store. The pmf of $X$ is
$\begin{array}{c|ccccc}{x} & {0} & {1} & {2} & {3} & {4} \\ \hline p_{X}(x) & {.1} & {.2} & {.3} & {.25} & {.15} \\ \hline\end{array}$
Sixty percent of all customers who purchase these cameras also buy an extended warranty. tell $Y$ denote the number of purchasers during this week who buy an extended warranty.
(a) What is $P(X=4, Y=2) ?[$Hint: This probability equals $P(Y=2 | X=4) \cdot P(X=4) ;$ now think of the four purchases as four trials of a binomial experiment, with success on a trial corresponding to buying an extended warranty 1
(b) Calculate $P(X=Y)$
(c) Determine the joint pmf of $X$ and $Y$ and then the marginal pmf of $Y .$

Key Concepts

Probability Mass Function (PMF)

A probability mass function defines the probabilities of occurrence of different outcomes for a discrete random variable. It maps each possible value of the variable to a probability, ensuring that the sum of these probabilities equals 1. This concept is fundamental in understanding how the probability is distributed over the outcomes of the random variable.

Conditional Probability

Conditional probability measures the probability of one event occurring given that another event has already occurred. In problems involving multiple random variables, it helps in computing joint probabilities by considering the dependency between the variables. This approach is essential when events are sequential or when one event influences the probability mechanism of the other.

Binomial Distribution

The binomial distribution describes the number of successes in a fixed number of independent Bernoulli trials, each having the same probability of success. It is applicable when there are two possible outcomes (success or failure) for each trial. This distribution is frequently used in problems where the probability of each trial’s success is known and the trials are independent.

Joint Distribution

The joint distribution of two random variables provides the probability that each of the variables falls within a specific range or takes a specific value simultaneously. It allows a comprehensive view of the probability structure of the two variables, including any dependency that exists between them, and is often used to perform further calculations involving multimodal events.

Marginal Distribution

The marginal distribution is obtained by summing (or integrating) the joint probability distribution over the range of the other variable(s). It gives the probabilities of a subset of the random variables irrespective of the values of the others, thereby offering a simplified view of the overall probability structure for individual random variables.

Transcript

00:01 The marginal distribution of x is given to us in the question.

00:04 So i will just rewrite that part.

00:06 So the marginal distribution of x.

00:12 So i'm going to draw that table here.

00:15 So they have given us the values of x as a 0.

00:21 Then we have 1.

00:23 Next is 2, 3, 4.

00:27 And the probability of each of these values is also known.

00:32 So which is 0 .1, next is 0 .2, then we have 0 .3, then 0 .25 and 0 .15.

00:44 Okay.

00:44 Now, to start with, they have told that x over here represents the number of canon digital cameras sold during a particular week by a certain store.

00:55 And y over here represents the number of purchases during this week who buy an extended warranty.

01:02 So to start with the first part, what they have asked in the question, we can say that n over here is given as 4.

01:09 And the probability where they have mentioned that 60 % of customers who purchase camera also by warranty, which means 60 % is 0 .6 over here as a probability.

01:20 So the conditional probability in this case, we can write what is asked where x is 4, depending on when y is 2.

01:40 This is given by probability of y equal to 2 when x is equal to 4 and this into probability of x is equal to 4 so over here we're going to use the binomial expansion formula so that formula is given by n c x then we have the probability raised to x and 1 minus p raised to n minus x so this is the formula that we're going to use for the binomial expansion so if i have a probability raise to x and 1 minus x so this is the formula i solve this part over here, then i can start writing the values that this will be equal to 4c2 and 0 .6 raised to 2.

02:23 Then we have 0 .4 raised to 4 minus 2 is 2.

02:27 And into the probability of x is equal to 4 is 0 .15.

02:32 So now when i substitute these values, then we get the value as 0 .05184.

02:40 So as i said, we have taken 4c2 because we have two yellow and two values of y and x is equal to 4.

02:51 So based on that, we have applied this formula here.

02:54 And moving on to the next part, they have told us to find the joint probability where p of x is equal to y.

03:05 So for the part b, they have told to find p of x is equal to y.

03:09 So this means that we have to consider all the values of x and y when they both have the same value where it will be from 0 up to 4 and we multiply that with the probability of that particular value.

03:22 So to start with the part b, i would say p of x is equal to y.

03:30 We can write this in detail as with the formula.

03:33 And p of y is 0, x is 0 and we take into probability of x is 0, right? plus the next value is when both are 1.

03:45 So p of y is 1 when x is 1 into p of x is 1.

03:53 Then i have plus p of y is 2 when x is 2 and into this is p of x is equal to 2.

04:04 Next we have p of y is 3 when x is 3 and into probability of x is equal to 3.

04:13 And lastly the value p of y is 4 when x is 4 and into probability of x is equal to 4 so the pmfs are already given to us also the binomial expansion formula that i apply so we know the value of p and n over there so i can directly apply and the formula for the first one is 0 c0 we will have 0 .6 raise to 0 then the 0 .4 value is also raised to 0 and into the p of x equal to 0 value is 0 .1.

04:50 Plus the next is 1c1, then we have 0 .6 raised to 1.

04:55 Again 0 .4 will be raised to 0 and into x equal to 1 probability is given as 0.

05:02 Further, the next value is 2c2.

05:05 We take 0 .6 raised to 2 .4 is again raised to 0 .4 into probability of x equal 2 is 0 .3...

Please give Ace some feedback