Question

an experiment consists of randomly selecting a single digit (from 0 to 9) where each digit is equally likely to be selected. define the following events: a = an even digit is selected b = a digit greater than 3 is selected c = a digit less than 7 is selected d = 1 or 7 selected which two events are independent? note: 0 is an even number. c and d a and c b and c a and b a and d b and d

Explanation:

Step1: Calculate number of elements in sample - space

The sample - space $S=\{0,1,2,3,4,5,6,7,8,9\}$, so $n(S) = 10$.

Step2: Calculate $n(A)$

The even digits are $0,2,4,6,8$, so $n(A)=5$ and $P(A)=\frac{n(A)}{n(S)}=\frac{5}{10}=0.5$.

Step3: Calculate $n(B)$

The digits greater than 3 are $4,5,6,7,8,9$, so $n(B)=6$ and $P(B)=\frac{n(B)}{n(S)}=\frac{6}{10}=0.6$.

Step4: Calculate $n(C)$

The digits less than 7 are $0,1,2,3,4,5,6$, so $n(C)=7$ and $P(C)=\frac{n(C)}{n(S)}=\frac{7}{10}=0.7$.

Step5: Calculate $n(D)$

The digits 1 or 7 are selected, so $n(D)=2$ and $P(D)=\frac{n(D)}{n(S)}=\frac{2}{10}=0.2$.

Step6: Calculate $n(A\cap B)$

The elements in $A\cap B$ are $4,6,8$, so $n(A\cap B)=3$ and $P(A\cap B)=\frac{n(A\cap B)}{n(S)}=\frac{3}{10}=0.3$. Since $P(A)\times P(B)=0.5\times0.6 = 0.3=P(A\cap B)$.

Step7: Check other intersections

• For $C$ and $D$: $n(C\cap D)=1$ (the digit 1), $P(C\cap D)=\frac{1}{10}=0.1$, $P(C)\times P(D)=0.7\times0.2 = 0.14

eq P(C\cap D)$.

• For $A$ and $C$: $n(A\cap C)=4$ (the digits $0,2,4,6$), $P(A\cap C)=\frac{4}{10}=0.4$, $P(A)\times P(C)=0.5\times0.7 = 0.35

eq P(A\cap C)$.

• For $B$ and $C$: $n(B\cap C)=3$ (the digits $4,5,6$), $P(B\cap C)=\frac{3}{10}=0.3$, $P(B)\times P(C)=0.6\times0.7 = 0.42

eq P(B\cap C)$.

• For $A$ and $D$: $n(A\cap D)=0$, $P(A\cap D)=0$, $P(A)\times P(D)=0.5\times0.2 = 0.1

eq P(A\cap D)$.

• For $B$ and $D$: $n(B\cap D)=1$ (the digit 7), $P(B\cap D)=\frac{1}{10}=0.1$, $P(B)\times P(D)=0.6\times0.2 = 0.12

eq P(B\cap D)$.

Answer:

A and B