Answer to Question #292293 in Statistics and Probability for Yaku

Question

5. A class in statistics contains 10 students, 3 of whom are 19, 4 are 20, 1

is 21, 1 is 24, and 1 is 26. Let X be the average age of the 2 randomly

selected students and derive the probability function for X.

6. A man has four keys in his pocket and, since it is dark, cannot see which

is his door key. He will try each key in turn until he finds the right one.

Let X be the number of keys tried (including the right one) to open the

door. What is the probability function for X?

7. Suppose a fair die is tossed two times. Let X be the larger of the two

faces that appear. Find _px(k).

8. Suppose a particle moves along the x-axis beginning at 0. It moves one

integer step to the left or right with equal probability. What is the probability

function of its position after four steps?

9. Five cards are dealt from a standard 52-card deck. Let Y be the number

of red cards that are dealt. What is the probability function for Y ?

Answer 1

5. We can choose $\dbinom{10}{2}=45$ pairs

$$(19, 19), (19, 19), (19, 20), (19, 20), (19, 20),$$ $$(19, 20),(19, 21), (19, 24), (19, 26), (19, 19),$$ $$(19, 20), (19, 20),(19, 20), (19, 20),(19, 21),$$ $$(19, 24),(19, 26), (19, 20),(19, 20),(19, 20),$$ $$(19, 20),(19, 21), (19, 24),(19, 26),(20, 20),$$ $$(20, 20), (20, 20), (20, 21), (20, 24), (20,26),$$ $$(20, 20), (20, 20), (20, 21), (20, 24), (20, 26),$$ $$(20, 20), (20, 21), (20, 24), (20, 26), (20, 21),$$ $$(20, 24), (20, 26), (21, 24), (21, 26), (24, 26)$$ $$\def\arraystretch{1.5} \begin{array}{c:c:c:c} & x_i & f_i & f(x_i) \\ \hline & 19 & 3& 1/15 \\ \hdashline & 19.5 & 12 & 4/15 \\ \hdashline & 20 & 9 & 3/15 \\ \hdashline & 20.5 & 4 & 4/45 \\ \hdashline & 21.5 & 3 & 1/15 \\ \hdashline & 22 & 4 & 4/45 \\ \hdashline & 22.5 & 4 & 4/45 \\ \hdashline & 23 & 4 & 4/45 \\ \hdashline & 23.5 & 1 & 1/45 \\ \hdashline & 25 & 1 & 1/45 \\ \hdashline \end{array}$$ $$\def\arraystretch{1.5} \begin{array}{c:c} x & p(x) \\ \hline 19 & 1/15 \\ \hdashline 19.5 & 4/15 \\ \hdashline 20 & 3/15 \\ \hdashline 20.5 & 4/45 \\ \hdashline 21.5 & 1/15 \\ \hdashline 22 & 1/15 \\ \hdashline 22.5 & 4/45 \\ \hdashline 23 & 4/45 \\ \hdashline 23.5 & 1/45 \\ \hdashline 25 & 1/45 \\ \hdashline \end{array}$$

6.

$$P(X=1)=\dfrac{1}{4}$$ $$P(X=2)=(1-\dfrac{1}{4})\dfrac{1}{4-1}=\dfrac{1}{4}$$ $$P(X=3)=(1-\dfrac{1}{4})(1-\dfrac{1}{4-1})\dfrac{1}{4-2}=\dfrac{1}{4}$$ $$P(X=4)=(1-\dfrac{1}{4})(1-\dfrac{1}{4-1})(1-\dfrac{1}{4-2})(1)=\dfrac{1}{4}$$ $$\def\arraystretch{1.5} \begin{array}{c:c} x & 1 & 2 & 3 & 4 \\ \hline p(x)& 1/4 & 1/4 & 1/4 & 1/4 \\ \end{array}$$

7. We have $6^2=36$ pairs

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c:c} & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline 1 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hdashline 2 & 2 & 2 & 3 & 4 & 5 & 6 \\ \hdashline 3 & 3 & 3 & 3 & 4 & 5 & 6 \\ \hdashline 4 & 4 & 4 & 4 & 4 & 5 & 6 \\ \hdashline 5 & 5 & 5 & 5 & 5 & 5 & 6 \\ \hdashline 6 & 6 & 6 & 6 & 6 & 6 & 6 \\ \hdashline \end{array}$$ $$\def\arraystretch{1.5} \begin{array}{c:c:c:c} & x_i & f_i & f(x_i) \\ \hline & 1 & 1 & 1/36 \\ \hdashline & 2 & 3 & 1/12 \\ \hdashline & 3 & 5 & 5/36 \\ \hdashline & 4 & 7 & 7/36 \\ \hdashline & 5 & 9 & 1/4 \\ \hdashline & 6 & 11 & 11/36 \\ \hdashline \end{array}$$ $$\def\arraystretch{1.5} \begin{array}{c:c} x & p(x) \\ \hline 1 & 1/36 \\ \hdashline 2& 1/12 \\ \hdashline 3 & 5/36 \\ \hdashline 4 & 7/36 \\ \hdashline 5 & 1/4 \\ \hdashline 6& 11/36 \\ \hdashline \end{array}$$

8.

$$\def\arraystretch{1.5} \begin{array}{c:c} x & p(x) \\ \hline -4 & 1/16 \\ \hdashline -2& 1/4\\ \hdashline 0 & 3/8 \\ \hdashline +2 & 1/4 \\ \hdashline +4& 1/16\\ \end{array}$$

9. There are 13 red cards in a deck

$$\dbinom{52}{5}=2598960$$ $$P(X=0)=\dfrac{\dbinom{13}{0}\dbinom{52-13}{5-0}}{\dbinom{52}{5}}=\dfrac{1(575757)}{2598960}$$ $$=0.2215336$$ $$P(X=1)=\dfrac{\dbinom{13}{1}\dbinom{52-13}{5-1}}{\dbinom{52}{5}}=\dfrac{13(82251)}{2598960}$$ $$=0.4114196$$ $$P(X=2)=\dfrac{\dbinom{13}{2}\dbinom{52-13}{5-2}}{\dbinom{52}{5}}=\dfrac{78(9139)}{2598960}$$ $$=0.2742797$$ $$P(X=3)=\dfrac{\dbinom{13}{3}\dbinom{52-13}{5-3}}{\dbinom{52}{5}}=\dfrac{286(741)}{2598960}$$ $$=0.0815426$$ $$P(X=4)=\dfrac{\dbinom{13}{4}\dbinom{52-13}{5-4}}{\dbinom{52}{5}}=\dfrac{715(39)}{2598960}$$ $$=0.0107293$$ $$P(X=5)=\dfrac{\dbinom{13}{5}\dbinom{52-13}{5-5}}{\dbinom{52}{5}}=\dfrac{1287(1)}{2598960}$$ $$=0.0004952$$ $$\def\arraystretch{1.5} \begin{array}{c:c} x & p(x) \\ \hline 1 & 0.4114196 \\ \hdashline 2& 0.2742797 \\ \hdashline 3 & 0.0815426 \\ \hdashline 4 & 0.0107293 \\ \hdashline 5 & 0.0004952 \\ \end{array}$$