Combined & Conditional Probability Flashcards

All 17 cards in this deck

What is the multiplication rule for two independent events?

e.g. $P(\text{Owen scores})=0.4$, $P(\text{Wasim scores})=0.25$

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

e.g. $P(\text{both score}) = 0.4 \times 0.25 = 0.1$

What is the addition rule for mutually exclusive events?

e.g. $P(\text{win})=0.3$, $P(\text{draw})=0.1$ in one game

$P(A \text{ or } B) = P(A) + P(B)$

e.g. $P(\text{win or draw}) = 0.3 + 0.1 = 0.4$

True or false? If $P(\text{Owen scores}) = 0.4$ and $P(\text{Wasim scores}) = 0.25$, then the probability that both score is $0.4 + 0.25$.

False. For 'both', independent probabilities should be multiplied: $0.4 \times 0.25$.

True or false? $P(A \text{ or } B) = P(A) + P(B)$ for any two events $A$ and $B$.

False. Adding only works for mutually exclusive events; otherwise an outcome could be in both categories and gets counted twice.

How do you find the probability of 'at least one' of several events happening?

e.g. two tests, $P(\text{fails both}) = 0.0351$

$1 - P(\text{none of them happen})$

e.g. $P(\text{passes at least one}) = 1 - 0.0351 = 0.9649$

What are the steps to find the probability that exactly one of two independent events happens?

e.g. $P(A)=0.3$, $P(B)=0.5$

1. $P(A \text{ and not } B) = 0.3 \times 0.5 = 0.15$
2. $P(\text{not } A \text{ and } B) = 0.7 \times 0.5 = 0.35$
3. Add the two: $0.15 + 0.35 = 0.5$

What does the notation $P(B \mid A)$ mean?

The probability of $B$ given that $A$ has happened.

In a conditional probability such as 'given that the customer uses type A', what must the denominator be?

The total number in the given group only (all those who use type A), not the overall total.

What are the steps to find a conditional probability from a two-way table?

e.g. one of the 22 females is chosen; find $P(\text{said Spain})$, where 3 females said Spain

1. Find the total of the given row or column: 22 females
2. Find how many of those are in the wanted category: 3
3. Divide: $\frac{3}{22}$

What are the steps to find $P(B \mid A)$ from a Venn diagram?

e.g. 12 members in set $A$ altogether, 4 of them also in $B$

1. Restrict the sample space to set $A$: 12
2. Count those in $A$ that are also in $B$: 4
3. Divide: $\frac{4}{12} = \frac{1}{3}$

True or false? $P(A \mid B)$ is always equal to $P(B \mid A)$.

False. They have different denominators, so they are usually different.

True or false? If $P(A \mid B) = P(A)$, then $A$ and $B$ are independent.

True. Knowing $B$ has happened does not change the probability of $A$.

When an object is taken without replacement, what happens to the probabilities for the second selection?

The total goes down by 1, and the number of that colour goes down by 1 if one was taken.

e.g. 4 red of 12, first red taken: $P(\text{second red}) = \frac{3}{11}$

True or false? When two counters are taken from a bag without replacement, the two selections are independent events.

False. They are dependent — the first counter changes what is left for the second.

What are the steps to find the probability that 3 objects taken without replacement are all the same colour?

e.g. 4 red counters in a bag of 12, find $P(3 \text{ reds})$

1. First pick: $\frac{4}{12}$
2. Reduce top and bottom by 1 each time: $\frac{3}{11}$, then $\frac{2}{10}$
3. Multiply: $\frac{4}{12}\times\frac{3}{11}\times\frac{2}{10} = \frac{1}{55}$

What are the steps to find the probability of one of each colour when 2 objects are taken without replacement?

e.g. 3 red and 2 blue counters, find $P(\text{one of each})$

1. $P(\text{red then blue}) = \frac{3}{5}\times\frac{2}{4} = \frac{6}{20}$
2. $P(\text{blue then red}) = \frac{2}{5}\times\frac{3}{4} = \frac{6}{20}$
3. Add both orders: $\frac{12}{20} = \frac{3}{5}$

In a bag of $n$ counters of which $r$ are red, what is the probability that the first two taken without replacement are both red?

$\frac{r}{n}\times\frac{r-1}{n-1}$