# Using Probability

1.** $\dfrac{1}{36}$.** The probability of getting a 5 on the first throw is $\dfrac{1}{6}$ and the probability of getting a 5 on the second throw is $\dfrac{1}{6}$. To find the probability of both happening, you multiply: $\dfrac{1}{6} \times \dfrac{1}{6} = \dfrac{1}{36}$.

2.** $\dfrac{1}{18}$.** The probability of getting a 6 or a 1 on the first throw is $\dfrac{2}{6}$ and the probability of getting the other number (either a 6 or a 1, depending on what you got on the first throw) on the second throw is $\dfrac{1}{6}$. To find the probability of both happening, you multiply: $\dfrac{2}{6} \times \dfrac{1}{6} = \dfrac{2}{36}=\dfrac{1}{18}$.

3.** $\dfrac{1}{36}$.** There is only one way for the sum of the rolls to be 12: if both rolls are a 6. The probability of getting a 6 on the first throw is $\dfrac{1}{6}$ and the probability of getting a 6 on the second throw is $\dfrac{1}{6}$. To find the probability of both happening, you multiply: $\dfrac{1}{6} \times \dfrac{1}{6} = \dfrac{1}{36}$.

4.** $\dfrac{1}{22}$.** The probability of getting a blue sock on the first pull is $\dfrac{3}{12}$ and the probability of getting a blue sock on the second pull $\dfrac{2}{11}$ (because you have already removed one blue sock from the drawer). To find the probability of both happening, you multiply: $\dfrac{3}{12} \times \dfrac{2}{11} = \dfrac{6}{132}=\dfrac{3}{66}=\dfrac{1}{22}$.

5.** $\dfrac{1}{11}$.** The probability of getting a white sock on the first pull is $\dfrac{4}{12}$ and the probability of getting a white sock on the second pull $\dfrac{3}{11}$ (because you have already removed one white sock from the drawer). To find the probability of both happening, you multiply: $\dfrac{4}{12} \times \dfrac{3}{11} = \dfrac{12}{132}=\dfrac{6}{66}=\dfrac{1}{11}$.

6. **$\dfrac{19}{66}$.** To solve this problem you have to find the odds of getting any of three pairs. You already found that the odds of getting a blue pair is $\dfrac{1}{22}$. The odds of getting a white pair is $\dfrac{1}{11}$. Now the black pair: the probability of getting a black sock on the first pull is $\dfrac{5}{12}$ and the probability of getting a black sock on the second pull $\dfrac{4}{11}$ (because you have already removed one black sock from the drawer). To find the probability of both happening, you multiply: $\dfrac{5}{12} \times \dfrac{4}{11} = \dfrac{20}{132}=\dfrac{10}{66}=\dfrac{5}{33}$. Now to find the odds of ANY of those pairs coming out, you add those probabilities (let's put them back into the same denominator): $\dfrac{6}{132} + \dfrac{12}{132} + \dfrac{20}{132} = \dfrac{38}{132}= \dfrac{19}{66}$.

7. **$\dfrac{1}{52} + \dfrac{1}{51} + \dfrac{1}{50} + \dfrac{1}{49} + \dfrac{1}{48} =$.** The probability of getting a queen of hearts on the first pull is $\dfrac{1}{52}$, on the second pull $\dfrac{1}{51}$, on the third pull $\dfrac{1}{50}$, on the fourth pull $\dfrac{1}{49}$, on the fifth pull $\dfrac{1}{48}$. To find the probability of it ever happening, you add those probabilities: $\dfrac{1}{52} + \dfrac{1}{51} + \dfrac{1}{50} + \dfrac{1}{49} + \dfrac{1}{48} =$.

8. **$\dfrac{4}{52}$.** The probability of getting a queen is $\dfrac{4}{52}$.

9. **$\dfrac{1}{52} + \dfrac{1}{51} + \dfrac{1}{50} + \dfrac{1}{49} + \dfrac{1}{48} =$.** The probability of getting a queen on the first pull is $\dfrac{4}{52}$, and the probability of getting an ace on the second pull $\dfrac{4}{51}$, To find the probability of both happening, you multiply those probabilities: $\dfrac{4}{52} \times \dfrac{4}{51} = \dfrac{16}{2652} = \dfrac{4}{663}$.

10. There could be **11** kids in the class. There could also be **22** (if two stars are chosen each week). Actually, there could be any multiple of 11 in the class (33, 44, 55, etc). The bottom of the probability is the total, but remember, fractions can be reduced, so any multiple of 11 could be the total.