Combined Conditional Probabilities Flashcards

All 5 cards in this deck

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?

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