Question

digits in exercises 6 - 9, each of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 is written on a piece of paper and all the pieces of paper are placed in a hat. one number is randomly selected. determine the probability that the number selected is 6. even. 7. odd or greater than 3. 8. greater than 2 or less than 6. 9. even and greater than 4.

Explanation:

Step1: Determine total number of outcomes

There are 10 digits (0 - 9), so the total number of possible outcomes $n(S)=10$.

Step2: Solve for question 6

The even - digits are 0, 2, 4, 6, 8. So the number of favorable outcomes $n(A)=5$. The probability $P(A)=\frac{n(A)}{n(S)}=\frac{5}{10}=\frac{1}{2}$.

Step3: Solve for question 7

The odd digits are 1, 3, 5, 7, 9. The digits greater than 3 are 4, 5, 6, 7, 8, 9. The set of odd or greater than 3: $\{1,3,4,5,6,7,8,9\}$, so $n(B)=8$. The probability $P(B)=\frac{n(B)}{n(S)}=\frac{8}{10}=\frac{4}{5}$.

Step4: Solve for question 8

The digits greater than 2 are 3, 4, 5, 6, 7, 8, 9. The digits less than 6 are 0, 1, 2, 3, 4, 5. The set of greater than 2 or less than 6: $\{0,1,2,3,4,5,6,7,8,9\}$, so $n(C)=10$. The probability $P(C)=\frac{n(C)}{n(S)} = 1$.

Step5: Solve for question 9

The even digits greater than 4 are 6, 8. So $n(D)=2$. The probability $P(D)=\frac{n(D)}{n(S)}=\frac{2}{10}=\frac{1}{5}$.

Answer:

1. $\frac{1}{2}$
2. $\frac{4}{5}$
3. 1
4. $\frac{1}{5}$