Choose the event before choosing the rule

1. 1
   Name the complement

   Not late is the complement of L.

   $$P(L^c)=1-P(L)$$
2. 2
   Subtract

   Remove the late probability from the whole sample space.

   $$1-0.63=0.37$$

Independent check: The two exhaustive probabilities 0.63 and 0.37 add to 1.

1. 1
   Choose inclusion-exclusion

   The word or asks for a union.

   $$P(F\cup M)=P(F)+P(M)-P(F\cap M)$$
2. 2
   Substitute

   Subtract the duplicated overlap.

   $$0.48+0.37-0.15=0.70$$

Independent check: The disjoint regions F-only 0.33, both 0.15, and M-only 0.22 total 0.70.

1. 1
   Use independence

   Passing A does not change the chance of passing B.

   $$P(A\cap B)=P(A)P(B)$$
2. 2
   Multiply

   Combine the independent probabilities.

   $$(0.70)(0.40)=0.28$$

Independent check: A 10-by-10 equally likely grid has 70 A outcomes; 40 percent of those, or 28 cells, also satisfy B.

1. 1
   Identify the condition

   Weekly attendance is the conditioning event.

   $$P(V\mid W)=\frac{P(V\cap W)}{P(W)}$$
2. 2
   Divide

   Normalize the shared probability within W.

   $$0.12/0.30=0.40$$

Independent check: Multiplying 0.40 by the conditioning probability 0.30 recovers the intersection 0.12.

1. 1
   First draw

   Four of the 52 cards are aces.

   $$P(A_1)=\frac4{52}$$
2. 2
   Condition the second draw

   Given an ace first, three aces remain among 51 cards.

   $$P(A_2\mid A_1)=\frac3{51}$$
3. 3
   Multiply and reduce

   Use the general multiplication rule.

   $$\frac4{52}\frac3{51}=\frac1{221}$$

Independent check: Counting pairs gives 6 ace pairs among 1,326 total two-card combinations, and 6/1,326 reduces to 1/221.

1. 1
   Find A only

   Independence also applies to the complement of B.

   $$P(A\cap B^c)=(0.20)(0.50)=0.10$$
2. 2
   Find B only

   Use the complement of A.

   $$P(A^c\cap B)=(0.80)(0.50)=0.40$$
3. 3
   Add disjoint cases

   Exactly one consists of the two nonoverlapping regions.

   $$0.10+0.40=0.50$$

Independent check: Inclusion-exclusion gives 0.20 + 0.50 - 2(0.10) = 0.50.

1. 1
   Find no failures

   All three independent sensors must work.

   $$P(0\text{ fail})=(0.90)^3=0.729$$
2. 2
   Take the complement

   Every other outcome has at least one failure.

   $$1-0.729=0.271$$

Independent check: Adding the binomial probabilities for one, two, and three failures gives 0.243 + 0.027 + 0.001 = 0.271.

1. 1
   Count the sample space

   Each die has six outcomes.

   $$|S|=6\cdot6=36$$
2. 2
   Count favorable pairs

   There are five ordered pairs with sum 8.

   $$|E|=5$$
3. 3
   Form the ratio

   The 36 ordered outcomes are equally likely.

   $$P(E)=\frac5{36}$$

Independent check: The diagonal counts for sums 2 through 12 are 1,2,3,4,5,6,5,4,3,2,1; the sum-8 diagonal has five outcomes.

1. 1
   Check the model

   There are four independent, constant-probability, two-outcome trials.

   $$n=4,\quad k=2,\quad p=0.30$$
2. 2
   Apply the formula

   Count the six placements of two successes.

   $$\binom42(0.30)^2(0.70)^2$$
3. 3
   Evaluate

   Multiply without rounding intermediate values.

   $$6(0.09)(0.49)=0.2646$$

Independent check: Enumerating the six strings SSFF, SFSF, SFFS, FSSF, FSFS, and FFSS gives the same total.

1. 1
   Multiply the marginals

   Independence predicts this intersection.

   $$P(A)P(B)=(0.50)(0.40)=0.20$$
2. 2
   Compare

   The prediction equals the supplied intersection.

   $$P(A\cap B)=P(A)P(B)$$

Independent check: The conditional probability is 0.20/0.40 = 0.50, unchanged from P(A).