Probability worksheet

1. P(A and B), both happen: 0

1. Mutually exclusive events can never happen together, so the chance of both is zero

   $\text{P(A and B)}$$=0$

2. For at least one, add the two chances. Then take away the chance of both, so it isn't counted twice

   $\text{P(A or B)}$$=\text{P(A)}+\text{P(B)}-\text{P(A and B)}$$=0.5+0.3-0$$=0.8$

3. For A not happening, take its chance away from 1, because an event either happens or it doesn't

   $\text{P(not A)}$$=1-0.5$$=0.5$

4. For exactly one, start with at least one and take away the cases where both happen

   $\text{P(A xor B)}$$=\text{P(A or B)}-\text{P(A and B)}$$=0.8-0$$=0.8$

5. For neither, take the chance of at least one away from 1. What's left is every case where nothing happens

   $\text{P(neither)}$$=1-0.8$$=0.2$

2. P(A and B), both happen: 0

1. Mutually exclusive events can never happen together, so the chance of both is zero

   $\text{P(A and B)}$$=0$

2. For at least one, add the two chances. Then take away the chance of both, so it isn't counted twice

   $\text{P(A or B)}$$=\text{P(A)}+\text{P(B)}-\text{P(A and B)}$$=0.4+0.2-0$$=0.6$

3. For A not happening, take its chance away from 1, because an event either happens or it doesn't

   $\text{P(not A)}$$=1-0.4$$=0.6$

4. For exactly one, start with at least one and take away the cases where both happen

   $\text{P(A xor B)}$$=\text{P(A or B)}-\text{P(A and B)}$$=0.6-0$$=0.6$

5. For neither, take the chance of at least one away from 1. What's left is every case where nothing happens

   $\text{P(neither)}$$=1-0.6$$=0.4$

3. P(A and B), both happen: 0.15

1. Multiply the two chances to get the chance both happen. This works because one event doesn't change the other's odds

   $\text{P(A and B)}$$=\text{P(A)}\times \text{P(B)}$$=0.5\times 0.3$$=0.15$

2. For at least one, add the two chances. Then take away the chance of both, so it isn't counted twice

   $\text{P(A or B)}$$=\text{P(A)}+\text{P(B)}-\text{P(A and B)}$$=0.5+0.3-0.15$$=0.65$

3. For A not happening, take its chance away from 1, because an event either happens or it doesn't

   $\text{P(not A)}$$=1-0.5$$=0.5$

4. For exactly one, start with at least one and take away the cases where both happen

   $\text{P(A xor B)}$$=\text{P(A or B)}-\text{P(A and B)}$$=0.65-0.15$$=0.5$

5. For neither, take the chance of at least one away from 1. What's left is every case where nothing happens

   $\text{P(neither)}$$=1-0.65$$=0.35$

4. P(A and B), both happen: 0.24

1. Multiply the two chances to get the chance both happen. This works because one event doesn't change the other's odds

   $\text{P(A and B)}$$=\text{P(A)}\times \text{P(B)}$$=0.6\times 0.4$$=0.24$

2. For at least one, add the two chances. Then take away the chance of both, so it isn't counted twice

   $\text{P(A or B)}$$=\text{P(A)}+\text{P(B)}-\text{P(A and B)}$$=0.6+0.4-0.24$$=0.76$

3. For A not happening, take its chance away from 1, because an event either happens or it doesn't

   $\text{P(not A)}$$=1-0.6$$=0.4$

4. For exactly one, start with at least one and take away the cases where both happen

   $\text{P(A xor B)}$$=\text{P(A or B)}-\text{P(A and B)}$$=0.76-0.24$$=0.52$

5. For neither, take the chance of at least one away from 1. What's left is every case where nothing happens

   $\text{P(neither)}$$=1-0.76$$=0.24$

5. P(A and B), both happen: 0.16

1. Multiply the two chances to get the chance both happen. This works because one event doesn't change the other's odds

   $\text{P(A and B)}$$=\text{P(A)}\times \text{P(B)}$$=0.4\times 0.4$$=0.16$

2. For at least one, add the two chances. Then take away the chance of both, so it isn't counted twice

   $\text{P(A or B)}$$=\text{P(A)}+\text{P(B)}-\text{P(A and B)}$$=0.4+0.4-0.16$$=0.64$

3. For A not happening, take its chance away from 1, because an event either happens or it doesn't

   $\text{P(not A)}$$=1-0.4$$=0.6$

4. For exactly one, start with at least one and take away the cases where both happen

   $\text{P(A xor B)}$$=\text{P(A or B)}-\text{P(A and B)}$$=0.64-0.16$$=0.48$

5. For neither, take the chance of at least one away from 1. What's left is every case where nothing happens

   $\text{P(neither)}$$=1-0.64$$=0.36$

6. P(A and B), both happen: 0.28

1. Multiply the two chances to get the chance both happen. This works because one event doesn't change the other's odds

   $\text{P(A and B)}$$=\text{P(A)}\times \text{P(B)}$$=0.7\times 0.4$$=0.28$

2. For at least one, add the two chances. Then take away the chance of both, so it isn't counted twice

   $\text{P(A or B)}$$=\text{P(A)}+\text{P(B)}-\text{P(A and B)}$$=0.7+0.4-0.28$$=0.82$

3. For A not happening, take its chance away from 1, because an event either happens or it doesn't

   $\text{P(not A)}$$=1-0.7$$=0.3$

4. For exactly one, start with at least one and take away the cases where both happen

   $\text{P(A xor B)}$$=\text{P(A or B)}-\text{P(A and B)}$$=0.82-0.28$$=0.54$

5. For neither, take the chance of at least one away from 1. What's left is every case where nothing happens

   $\text{P(neither)}$$=1-0.82$$=0.18$

7. P(A and B), both happen: 0.18

1. Multiply the two chances to get the chance both happen. This works because one event doesn't change the other's odds

   $\text{P(A and B)}$$=\text{P(A)}\times \text{P(B)}$$=0.6\times 0.3$$=0.18$

2. For at least one, add the two chances. Then take away the chance of both, so it isn't counted twice

   $\text{P(A or B)}$$=\text{P(A)}+\text{P(B)}-\text{P(A and B)}$$=0.6+0.3-0.18$$=0.72$

3. For A not happening, take its chance away from 1, because an event either happens or it doesn't

   $\text{P(not A)}$$=1-0.6$$=0.4$

4. For exactly one, start with at least one and take away the cases where both happen

   $\text{P(A xor B)}$$=\text{P(A or B)}-\text{P(A and B)}$$=0.72-0.18$$=0.54$

5. For neither, take the chance of at least one away from 1. What's left is every case where nothing happens

   $\text{P(neither)}$$=1-0.72$$=0.28$

8. P(A and B), both happen: 0

1. Mutually exclusive events can never happen together, so the chance of both is zero

   $\text{P(A and B)}$$=0$

2. For at least one, add the two chances. Then take away the chance of both, so it isn't counted twice

   $\text{P(A or B)}$$=\text{P(A)}+\text{P(B)}-\text{P(A and B)}$$=0.5+0.2-0$$=0.7$

3. For A not happening, take its chance away from 1, because an event either happens or it doesn't

   $\text{P(not A)}$$=1-0.5$$=0.5$

4. For exactly one, start with at least one and take away the cases where both happen

   $\text{P(A xor B)}$$=\text{P(A or B)}-\text{P(A and B)}$$=0.7-0$$=0.7$

5. For neither, take the chance of at least one away from 1. What's left is every case where nothing happens

   $\text{P(neither)}$$=1-0.7$$=0.3$

9. P(A and B), both happen: 0.1

1. Multiply the two chances to get the chance both happen. This works because one event doesn't change the other's odds

   $\text{P(A and B)}$$=\text{P(A)}\times \text{P(B)}$$=0.5\times 0.2$$=0.1$

2. For at least one, add the two chances. Then take away the chance of both, so it isn't counted twice

   $\text{P(A or B)}$$=\text{P(A)}+\text{P(B)}-\text{P(A and B)}$$=0.5+0.2-0.1$$=0.6$

3. For A not happening, take its chance away from 1, because an event either happens or it doesn't

   $\text{P(not A)}$$=1-0.5$$=0.5$

4. For exactly one, start with at least one and take away the cases where both happen

   $\text{P(A xor B)}$$=\text{P(A or B)}-\text{P(A and B)}$$=0.6-0.1$$=0.5$

5. For neither, take the chance of at least one away from 1. What's left is every case where nothing happens

   $\text{P(neither)}$$=1-0.6$$=0.4$

10. P(A and B), both happen: 0

1. Mutually exclusive events can never happen together, so the chance of both is zero

   $\text{P(A and B)}$$=0$

2. For at least one, add the two chances. Then take away the chance of both, so it isn't counted twice

   $\text{P(A or B)}$$=\text{P(A)}+\text{P(B)}-\text{P(A and B)}$$=0.4+0.4-0$$=0.8$

3. For A not happening, take its chance away from 1, because an event either happens or it doesn't

   $\text{P(not A)}$$=1-0.4$$=0.6$

4. For exactly one, start with at least one and take away the cases where both happen

   $\text{P(A xor B)}$$=\text{P(A or B)}-\text{P(A and B)}$$=0.8-0$$=0.8$

5. For neither, take the chance of at least one away from 1. What's left is every case where nothing happens

   $\text{P(neither)}$$=1-0.8$$=0.2$