Question

ten identical slips of paper each contain one number from one to ten, inclusive. the papers are put into a bag and then mixed around. which statements about the situation are true? check all that apply. p(6)=p(1) p(5) = 1/2 p(>10)=0 p(1 < x < 10)=100% s={1,2,3,4,5,6,7,8,9,10} if a ⊂ s;a could be {1,3,5,7,9}

Explanation:

Step1: Calculate probability of single - number events

The probability of picking a single number from 10 equally - likely outcomes is given by the formula $P(x)=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$. Since there is 1 slip with the number 6 and 1 slip with the number 1 out of 10 slips, $P(6)=\frac{1}{10}$ and $P(1)=\frac{1}{10}$, so $P(6) = P(1)$.

Step2: Analyze $P(5)$

There is 1 slip with the number 5 out of 10 slips, so $P(5)=\frac{1}{10}
eq\frac{1}{2}$.

Step3: Analyze $P(> 10)$

Since the numbers on the slips are from 1 to 10, there are no numbers greater than 10. So the number of favorable outcomes for the event of picking a number greater than 10 is 0. Then $P(>10)=\frac{0}{10}=0$.

Step4: Analyze $P(1 < x < 10)$

The numbers satisfying $1 < x < 10$ are 2, 3, 4, 5, 6, 7, 8, 9. There are 8 such numbers out of 10. So $P(1 < x < 10)=\frac{8}{10}=80\%
eq100\%$.

Step5: Define the sample space

The sample space S is the set of all possible outcomes. Here, the possible outcomes are the numbers 1 to 10, so $S = \{1,2,3,4,5,6,7,8,9,10\}$.

Step6: Analyze subset A

If $A\subset S$, A is a subset of S. The set $\{1,3,5,7,9\}$ consists of elements that are in the set S, so if $A\subset S$, A could be $\{1,3,5,7,9\}$.

Answer:

$P(6) = P(1)$, $P(>10)=0$, $S = \{1,2,3,4,5,6,7,8,9,10\}$, If $A\subset S$; A could be $\{1,3,5,7,9\}$