Probability flash cards

1. 1.State Bayes' theorem for finding $P(E_i|A)$ given events $E_1, E_2, \ldots, E_n$ forming a partition of the sample space.

   $P(E_i|A)=\dfrac{P(E_i)P(A|E_i)}{\sum_{j=1}^{n}P(E_j)P(A|E_j)}$

   Hint: Combine the total probability theorem with the definition of conditional probability.

2. 2.Define conditional probability $P(A|B)$ and give the multiplication theorem for two events.

   $P(A|B)=\dfrac{P(A\cap B)}{P(B)}$, $P(B)\neq0$; hence $P(A\cap B)=P(B)\cdot P(A|B)=P(A)\cdot P(B|A)$.

   Hint: It is the ratio of the joint probability to the conditioning event's probability.

3. 3.What is the condition for two events $A$ and $B$ to be independent, and how does it extend to three events (pairwise vs mutual independence)?

   $A$ and $B$ are independent if $P(A\cap B)=P(A)\cdot P(B)$. For mutual independence of $A,B,C$: all three pairwise conditions $P(A\cap B)=P(A)P(B)$, $P(B\cap C)=P(B)P(C)$, $P(A\cap C)=P(A)P(C)$ must hold, and $P(A\cap B\cap C)=P(A)P(B)P(C)$.

   Hint: Pairwise independence alone is not sufficient for mutual independence.

4. 4.For a random variable $X$ with probability distribution $P(X=x_i)=p_i$, give the formulas for mean $E(X)$ and variance $\text{Var}(X)$.

   $E(X)=\sum x_ip_i$ and $\text{Var}(X)=\sum x_i^2p_i-[E(X)]^2=E(X^2)-[E(X)]^2$.

   Hint: Variance equals the second moment minus the square of the mean.

5. 5.State the probability mass function of a Binomial distribution $B(n,p)$ and its mean and variance.

   $P(X=r)=\,^{n}C_r\,p^rq^{n-r}$, for $r=0,1,\ldots,n$, where $q=1-p$. Mean $=np$, Variance $=npq$.

   Hint: Applies to repeated independent Bernoulli trials with a constant success probability.

6. 6.Two fair dice are thrown together. Given that the sum of the numbers shown is even, find the probability that both dice show even numbers.

   Sum is even iff both dice are odd or both are even, giving $9+9=18$ equally likely outcomes out of $36$. Of these, both-even accounts for exactly $9$ outcomes. So $P(\text{both even}\mid\text{sum even})=\dfrac{9}{18}=\dfrac{1}{2}$.

   Hint: Split the 'sum even' event into its two disjoint parity cases first.

7. 7.Three fair coins are tossed simultaneously. Given that at least one head appears, find the probability that exactly two heads appear.

   Out of $8$ equally likely outcomes, 'at least one head' excludes only TTT, giving $7$ outcomes. Exactly two heads (HHT, HTH, THH) gives $3$ outcomes. Hence $P(\text{exactly two heads}\mid\text{at least one head})=\dfrac{3}{7}$.

   Hint: Restrict the sample space to the 7 outcomes satisfying the condition.

8. 8.A family has exactly two children (each equally likely to be a boy B or girl G, independently). Compare $P(\text{both boys}\mid\text{at least one boy})$ with $P(\text{both boys}\mid\text{the elder child is a boy})$.

   Sample space $\{BB,BG,GB,GG\}$, each with probability $\dfrac14$. 'At least one boy' excludes only GG, leaving $\{BB,BG,GB\}$, so $P(\text{both boys}\mid\text{at least one boy})=\dfrac{1}{3}$. 'Elder is a boy' restricts to $\{BB,BG\}$, so $P(\text{both boys}\mid\text{elder is a boy})=\dfrac{1}{2}$. The two conditioning events are different, so the answers genuinely differ: $\dfrac13$ vs $\dfrac12$.

   Hint: Write out all 4 equally likely outcomes and see which ones each condition removes.

9. 9.A box has 4 red and 6 black balls. Two balls are drawn one after another without replacement. Given the first ball drawn is red, find the probability that the second ball drawn is also red.

   After removing one red ball, $3$ red and $6$ black balls remain out of $9$ total. So $P(\text{second red}\mid\text{first red})=\dfrac{3}{9}=\dfrac{1}{3}$.

   Hint: Just update the composition of the box after the first draw.

10. 10.An urn has 5 red and 7 black balls. Two balls are drawn one after another without replacement, but you are only told (after both draws) that the second ball drawn was red. Find the probability that the first ball drawn was also red.

    By symmetry $P(\text{2nd red})=\dfrac{5}{12}$. Also $P(\text{1st red, 2nd red})=\dfrac{5}{12}\cdot\dfrac{4}{11}=\dfrac{5}{33}$. So $P(\text{1st red}\mid\text{2nd red})=\dfrac{5/33}{5/12}=\dfrac{4}{11}$. Notice this equals $P(\text{2nd red}\mid\text{1st red})=\dfrac{4}{11}$, an instance of the exchangeability of unordered draws.

    Hint: Conditioning on a later event to infer an earlier one — use Bayes, not just forward logic.

11. 11.Two fair dice are rolled. Given that the numbers shown on the two dice are different, find the probability that their sum is 6.

    Outcomes with different numbers: $36-6=30$ (removing the 6 doubles). Sum $=6$ with distinct dice: $(1,5),(5,1),(2,4),(4,2)$ — note $(3,3)$ is excluded since the dice must differ — giving $4$ outcomes. So $P=\dfrac{4}{30}=\dfrac{2}{15}$.

    Hint: Remove the doubles from both the numerator and denominator counts.

12. 12.An urn contains 5 white and 7 black balls. Two balls are drawn in succession without replacement. Find $P(\text{second ball is white}\mid\text{first ball is black})$.

    After removing one black ball, $5$ white and $6$ black balls remain, total $11$. So $P(\text{second white}\mid\text{first black})=\dfrac{5}{11}$.

    Hint: Only the composition after the first draw matters for the second draw's probability.

13. 13.An integer is chosen at random from 1 to 20. Given that it is a multiple of 3, find the probability that it is also a multiple of 4.

    Multiples of $3$ in $[1,20]$: $\{3,6,9,12,15,18\}$, a set of $6$ numbers. A number that is a multiple of both $3$ and $4$ must be a multiple of $12$; only $12$ qualifies. So $P=\dfrac{1}{6}$.

    Hint: A number divisible by both 3 and 4 must be divisible by their LCM, 12.

Open the interactive deck for the other 72 cards, with self-grading so the ones you keep missing come back.