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Question 2 NAT · 5.0 marks
There are two stores,
and
, which sell t-shirts on the market. The store
has stock of
red
and 2 black t-shirts and the store
has a stock of 2 red and
black t-shirts. One of the stores is
selected at random and two t-shirts are purchased without replacement from that store. Given
that both the t-shirts purchased are red, the probability that they were purchased from store
is
, find the value of
.
Published solution
**1. Write the likelihoods:** The two stores are equally likely and both have n+2 shirts. The chance of two red shirts from A is \(\binom n2/\binom{n+2}2\); from B it is \(\binom22/\binom{n+2}2\).
**2. Apply Bayes' rule:** Equal store priors and the common denominator cancel, so
\[P(A\mid RR)=\frac{\binom n2}{\binom n2+1}=\frac67.\]
**3. Solve:** 7 times binomial(n,2)=6 times (binomial(n,2)+1), giving binomial(n,2)=6. Hence n(n-1)=12, so (n-4)(n+3)=0. A stock count is nonnegative, so n=4.
**Answer: 4.**
Three independent draws are made from a standard deck of 52 playing cards, with replacement
and reshuffling between each draw. Which of the following statements is/are correct?
6406536625631.
The probability that the first card drawn is the King of Hearts is
.
6406536625632.
The probability that both the first and second cards drawn are red is
6406536625633.
The probability that the third card drawn is the Ace of Spades is
.
6406536625634.
The events "the first card drawn is a Queen" and "the second card drawn is a
Queen" are independent events.
Published solution
**1. Use replacement:** The deck is restored and reshuffled between draws, so every draw has the same 52-card distribution, independently of earlier results.
**2. Compute the listed probabilities:** The King of Hearts on the first draw has probability 1/52. Two successive red cards have probability (26/52) squared=1/4. The Ace of Spades on the third draw still has probability 1/52, not 1/50.
**3. Test the Queen events:** The first-draw Queen event and second-draw Queen event are independent because of replacement; their joint probability is (4/52) squared.
**Answer: A, D.**
A: **Correct:** One particular card has probability 1/52 on every restored-deck draw.
B: **Incorrect:** Both red cards occur with probability 1/4, not 1/2.
C: **Incorrect:** Replacement keeps 52 cards available; the third-draw probability is 1/52.
D: **Correct:** The draw outcomes are independent under the stated replacement and reshuffling.
**1. Apply De Morgan:** Being outside both A and B is the same event as the complement of their union: (A union B) complement=A complement intersection B complement.
**2. Simplify the empty-set cases:** A union empty set=A, whereas A intersection empty set=empty set. Their probabilities are therefore P(A) and 0 respectively.
**3. Expand A complement union B:** Split it into all outcomes outside A, plus the disjoint additional part A intersection B. Its probability is 1-P(A)+P(A intersection B).
**Answer: A, C, D.**
A: **Correct:** This is exactly De Morgan's event identity.
B: **Incorrect:** The left side is P(A), not generally P(A complement).
C: **Correct:** The decomposition into disjoint A complement and A intersection B gives this formula.
D: **Correct:** The intersection with the empty set is empty and has probability zero.
A person won a special pass to an amusement park that allows him to bring 4 friends with him for
free. He has 8 friends, but two of them are twins who will only attend if both are included. How
many different groups of 4 friends can the person choose to take along?
Published solution
**1. Separate the twin cases:** Either both twins are chosen or neither is chosen. Choosing exactly one twin is not allowed.
**2. Count both included:** The twins fill two of the four places. Choose two of the other six friends: binomial(6,2)=15.
**3. Count neither included:** Choose all four friends from the other six: binomial(6,4)=15.
**4. Add disjoint possibilities:** 15+15=30 groups. These are groups, not ordered sequences, so combinations are appropriate.
**Answer: 30.**
Nine friends are to be seated around a round table. Among them, Amit and Rohit insist on sitting
next to each other. In how many ways can the friends be seated around a circular table?
Published solution
**1. Create an adjacent block:** Treat Amit and Rohit as one object. Alongside the other seven friends this produces eight objects to arrange around the circle.
**2. Account for circular symmetry:** Eight distinct objects have (8-1)!=7!=5,040 circular orders when rotations are equivalent.
**3. Order the pair:** Amit-Rohit and Rohit-Amit are two distinct internal orders. Thus total=2 times 7!=10,080. Reflections are not identified unless the problem explicitly says so.
**Answer: 10080.**
**1. Expand the combinations:**
\[\binom n2+2\binom n1=\frac{n(n-1)}2+2n=35.\]
**2. Simplify:** Multiply by 2 to get n squared-n+4n=70, or n squared+3n-70=0.
**3. Factor:** (n+10)(n-7)=0. The combination parameter must be a nonnegative integer with n at least 2, so n=7. Check: binomial(7,2)+2 binomial(7,1)=21+14=35.
**Answer: 7.**
**1. Read the proportions:** Let C mean cake, B mean bread, and S mean defective. P(C)=0.35, P(B)=0.40 and P(S)=0.06. The defect probability is the same for every item type, so defect status and type are independent.
**2. Find cake and defect:** P(C intersection S)=0.35 times 0.06=0.021. Also P(S given C)=0.06 and P(C given S)=0.35.
**3. Compare posterior types:** Because defect risk is equal across types, seeing a defective item does not change their proportions. Hence P(B given S)=0.40 is larger than P(C given S)=0.35.
**Answer: A, D.**
A: **Correct:** The independent joint probability is 0.35 times 0.06=0.021.
B: **Incorrect:** The given cake defect rate is 0.06, not 0.10.
C: **Incorrect:** This reverses the conditioning; P(C given S)=0.35, not 0.06.
D: **Correct:** Equal defect rates preserve the original 0.40 versus 0.35 type proportions.
Amit wants to create a 4-digit numerical PIN for his ATM card. In how many ways can he create a
4-digit number between 5000 and 9000 using the digits 1, 3, 5, 6, 8, 9, if repetition of digits is
allowed?
Published solution
**1. Restrict the thousands digit:** To lie between 5,000 and 9,000 using the supplied digits, the first digit can be 5, 6 or 8. A first digit 9 would exceed 9,000 because no zero is available to make exactly 9,000.
**2. Fill the other places:** Repetition is allowed, so each of the remaining three positions has all six choices 1,3,5,6,8,9.
**3. Multiply:** 3 times 6 cubed=3 times 216=648 valid PINs.
**Answer: 648.**
**1. Interpret the count:** The question counts occurrences of the digit 5, so a number such as 555 contributes three occurrences, not one number.
**2. Use three-place representations:** Write 000 through 999 with leading zeros. In each of the hundreds, tens and units positions, the digit 5 occurs exactly 100 times because the other two positions have 10 choices each.
**3. Add positions:** Total occurrences=100+100+100=300. Removing 000 changes nothing and adding 1000 also adds no 5s. Counting numbers containing at least one 5 would instead give 1,000-9 cubed=271, which is a different question.
**Answer: B.**
A: **Incorrect:** 271 counts numbers containing a 5 at least once, not the total occurrences of the digit.
B: **Correct:** Each of three positions contributes 100 occurrences, giving 300.
C: **Incorrect:** The endpoints removed or added contain no 5, so no subtraction of one is needed.
D: **Incorrect:** The number 1000 contains no 5 and adds no occurrence.
From a group of 5 students, a team of 3 students is to be selected. From this selected team, one
student is chosen as the leader. In how many different ways can this be done?
6406536625646.
15
6406536625647.
20
6406536625648.
30
6406536625649.
60
Published solution
**1. Select the team:** Choosing three of five students gives binomial(5,3)=10 unordered teams.
**2. Assign the special role:** Any of the three selected students can be leader, giving three choices for each team.
**3. Multiply:** 10 times 3=30. As a check, choose the leader first in five ways, then the other two members from four students in binomial(4,2)=6 ways: 5 times 6=30.
**Answer: C.**
A: **Incorrect:** This omits half the possible team-and-leader assignments.
B: **Incorrect:** This does not correctly combine team selection with three possible leaders.
C: **Correct:** The team-and-role multiplication gives 30.
D: **Incorrect:** 60 orders all three members even though only the leader role distinguishes them.
A committee of 3 students is to be formed from a class of 7 boys and 5 girls.
Based on the given information, answer the given sub-questions.
In how many ways can a committee be formed if at least one girl must be included?
6406536625655.
140
6406536625656.
165
6406536625657.
185
6406536625658.
220
Published solution
**1. Count all committees:** The class has 12 students, so there are binomial(12,3)=220 three-student committees.
**2. Count the forbidden case:** No girls means selecting all three from the seven boys: binomial(7,3)=35.
**3. Take the complement:** At least one girl gives 220-35=185 committees. Subtracting all-boy committees automatically includes cases with one, two or three girls.
**Answer: C.**
A: **Incorrect:** This is not the total of all allowed one-, two- and three-girl cases.
B: **Incorrect:** This does not equal the required complement count.
C: **Correct:** Subtracting 35 all-boy committees from 220 gives 185.
D: **Incorrect:** 220 includes the forbidden all-boy committees.
**1. Apply the condition:** At least one boy excludes only all-girl committees. The conditioned sample space has binomial(12,3)-binomial(5,3)=220-10=210 committees.
**2. Count exactly one girl:** Choose one of five girls and two of seven boys: binomial(5,1) times binomial(7,2)=5 times 21=105.
**3. Divide:** Every favorable committee includes boys, so the conditional probability is 105/210=1/2. The denominator must reflect the stated condition, not all 220 committees.
**Answer: B.**
A: **Incorrect:** The correct favorable-to-conditioned count is 105/210, not 1/3.
B: **Correct:** 105 favorable committees out of 210 conditioned committees gives 1/2.
C: **Incorrect:** 5/12 refers to the marginal proportion of girls, not this committee event.
D: **Incorrect:** 7/10 is not the conditional ratio obtained from the committee counts.
Based on the above data, answer the given sub-questions.
Assuming
, find the value of
. Enter the answer correct to two
decimal places.
Published solution
**1. Apply the stated inclusion-exclusion calculation:**
\[P(A\cup B\cup C)=P(A)+P(B)+P(C)-P(A\cap B)-P(B\cap C)-P(A\cap C)+P(A\cap B\cap C).\]
Substituting the question's numbers and its explicit triple-intersection assumption gives 0.6+0.5+0.4-0.3-0.2-0.25+0=0.75.
**2. Check the model, not just the arithmetic:** These inputs are not jointly consistent as actual event probabilities when the triple intersection is zero. In particular,
\[P(C\setminus(A\cup B))=0.4-0.25-0.2+0=-0.05,\]
which is impossible.
**Interpretation:** 0.75 is the formal inclusion-exclusion result expected by the source, but the stated probability model needs correction. The calculation must not be taken as proof that such events exist.
**Answer: 0.75.**
