Ingrid and Thomas both enter a raffle where they get a chance to pull one winning ticket from a box. If there are 25 tickets in the box and tickets are not replaced, what is the probability that both will pull a losing ticket from the box? Write your answer as a simplified fraction.
Since the tickets are not replaced, after Ingrid pulls out a ticket, there will be 24 tickets left in the box. The probability that Ingrid pulls out a losing ticket is $\frac{24}{25}$. Then, after Thomas pulls out a ticket, there will be 23 tickets left in the box. The probability that Thomas pulls out a losing ticket is $\frac{23}{24}$. Therefore, the probability that both will pull a losing ticket is $\frac{24}{25}\times\frac{23}{24}=\boxed{\frac{23}{25}}$.
In a bag with plastic shapes, there are 6 triangles, 4 circles, and 5 squares. What is the probability of pulling out two triangles if the first shape is replaced before the second shape is pulled out? Write your answer as a simplified fraction.
Since the first triangle is replaced before the second shape is pulled out, the probability of pulling out a triangle on the first draw is $\frac{6}{6+4+5}=\frac{6}{15}$. The probability of pulling out a triangle on the second draw, assuming the first shape was replaced, is also $\frac{6}{15}$. Therefore, the probability of pulling out two triangles is $\left(\frac{6}{15}\right)\left(\frac{6}{15}\right)=\frac{36}{225}=\boxed{\frac{4}{25}}$.
How many outcomes are in the sample space of rolling two number cubes labeled 1–6? Use the method of your choice.
To find the number of outcomes in the sample space of rolling two number cubes labeled 1-6, we can use the counting principle. Since each cube has 6 faces, there are 6 possible outcomes for each cube. To find the total number of outcomes when rolling two cubes, we multiply the number of outcomes for each cube together. Therefore, there are $6 \times 6 = \boxed{36}$ outcomes in the sample space.
