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Let Xand Ybe the random variables that count the number of heads and the number of tails that come up when two fair coins are flipped. Click and drag statements to show that Xand Y are not independent Therefore, X and Y are not independentHowever, for r2=1 and r2=1, we have p(X-1 and Y-1)- 1/2 while p(x=1) : p(T=1)= (1/2):(1/2)=1However; for r2=2 and r2=2, we have p(X-2 and Y-2)-0 because X+Y must always be 2, while p(X=2) - p (Y=2) (1/4x1/4)=1/16X and Y are independent if p (X) - r1 and Y - r2) - p (X-r1), p (Y=r2) for all real nubers r1 and r2.X and Y are independent if p (X) - r1 and Y - r2) - p (X-r2), p (Y=r2) for all real nubers r1 and r2.

Sagot :

A variable whose values are determined by chance is called a random variable. Simply assigning different values to each occurrence, such as 0 for heads and 1 for tails, can be used to define a random variable for the two outcomes of a coin toss: "Heads" and "Tails."

In the event that a coin is thrown twice, there are 4 potential results: { TT, HH, HT, and TH

We can determine the values of X and Y for each of the four outcomes if X is a random variable that represents the number of tails in two tosses (either 1, 2, or 3) and Y is a second random variable that represents the number of heads on the second flip of the coin (either heads (Y=1) or tails (Y=0).

HH, for instance, is X=0, Y=1. We know X=0 on the grounds that neither one nor the other throws was tails, so the quantity of tails is 0. Because the second toss resulted in heads, Y=1.

Adding the remaining X and Y values for the remaining three outcomes:

HH: X=0, Y=1

HT: X=1, Y=0

TH: X=1, Y=1

TT: We are asked to determine the probability that X=1, Y=0. The probability of X=1, Y=0 is simply the number of events with that pair of values divided by the total number of possible outcomes because each of the four outcomes is equally likely to occur. X and Y are equal only in HT. As a result, P(X=1, Y=0) equals 1/4.

To learn more about random variable here

https://brainly.com/question/17238189

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