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Question 2 MCQ · 3.0 marks
In how many ways can the letters of the word "MATHEMATICS" be arranged such that all the
vowels appear together?
6406537087672.
60480
6406537087673.
30240
6406537087674.
241920
6406537087675.
120960
Published solution
**1. Group the vowels:** MATHEMATICS has 11 letters. Its vowels are A,A,E,I. Treat these four vowels as one block so that they stay together.
**2. Arrange the outer objects:** The seven consonants M,M,T,T,H,C,S plus the block make eight objects. Because M and T each repeat twice, their distinct arrangements number \(8!/(2!2!)=10,080\).
**3. Arrange within the block:** A,A,E,I can be arranged in \(4!/2!=12\) ways.
**4. Multiply:** \(10,080\times12=120,960\). Only repeated identical letters require factorial division.
**Answer: D.**
A: **Incorrect:** This is half the correct product of outer and internal arrangements.
B: **Incorrect:** This is one-quarter of the correct count.
C: **Incorrect:** This double-counts the two identical A vowels.
D: **Correct:** The block method with all repeated-letter corrections gives 120,960.
, are to be arranged around a circular
table.
Find the number of distinct seating arrangements if
and
sit together, but
does not sit next
to either
or
.
6406537087676.
720
6406537087677.
1440
6406537087678.
480
6406537087679.
960
Published solution
**1. Form the adjacent pair:** Treat A and B as a block. Together with C and the five others, this gives seven distinct objects around a circle. Their circular orders number (7-1)!=720, and the pair has two internal orders, giving 1,440.
**2. Remove forbidden neighbors:** C must not be adjacent to either end of the AB block. Among the six positions relative to the fixed block, two put C next to it. Equivalently the forbidden count is \(2\times5!\times2=480\): two sides for C and two internal orders of A,B.
**3. Subtract:** Valid arrangements=1,440-480=960. Rotations are identified in the circular count; reflected arrangements remain distinct.
**Answer: D.**
A: **Incorrect:** This counts the seven circular objects without the internal ordering and exclusion adjustments.
B: **Incorrect:** This keeps all AB-adjacent arrangements without excluding C next to the pair.
C: **Incorrect:** This is the forbidden count, not the desired count.
D: **Correct:** Subtracting the 480 forbidden arrangements from 1,440 gives 960.
A committee of 3 members is selected at random from 5 men and 4 women. Given that at least
one woman is selected, what is the probability that exactly one woman is selected?
**1. Restrict the sample space:** Total three-person committees number \(\binom93=84\). All-male committees number \(\binom53=10\). Given at least one woman, the possible committees therefore number 84-10=74.
**2. Count exactly one woman:** Choose one of four women and two of five men: \(\binom41\binom52=4\times10=40\).
**3. Form the conditional probability:** Every exactly-one-woman committee meets the condition, so the probability is 40/74=20/37. The denominator is not 84 because of the conditioning.
**Answer: C.**
A: **Incorrect:** 40/84 uses the unrestricted sample space and ignores the given condition.
B: **Incorrect:** 4/9 is the probability for one individual draw being female, not this committee event.
C: **Correct:** 40 favorable committees out of 74 conditioned committees gives 20/37.
D: **Incorrect:** 10/21 equals 40/84 and likewise omits conditioning.
A mobile phone lock uses a 4-character code. The first two characters are letters selected from
and
, where repetition of letters is allowed. The last two characters are digits selected
from 1, 2, 3, 4, and 5, where repetition of digits is not allowed. How many different lock codes can
be formed?
Published solution
**1. Count the letters:** The first and second positions each have four choices, A through D. Repetition is allowed, so there are 4 times 4=16 letter pairs.
**2. Count the digits:** There are five choices for the first digit and four remaining choices for the second because digits may not repeat. Thus 5 times 4=20 digit pairs.
**3. Apply the multiplication rule:** Each letter pair can be combined with each digit pair, giving 16 times 20=320 distinct four-character codes. No factorial division is needed because code positions are ordered.
**Answer: 320.**
Ten friends are seated randomly around a circular table. Two particular friends, Rahul and Priya,
wish to sit next to each other. Find the probability that Rahul and Priya are seated adjacent to each
other. Enter your answer correct to two decimal places.
Published solution
**1. Fix one person's seat:** Circular symmetry allows fixing Rahul. Priya is equally likely to occupy any of the other nine seats.
**2. Count adjacent positions:** Exactly two seats are next to Rahul, one on either side. Thus the probability is 2/9=0.222222... .
**3. Round:** To two decimal places, the answer is 0.22. The denominator is nine rather than ten because Rahul already occupies one seat.
**Answer: 0.22.**
A bag contains 5 red balls and 7 blue balls. Two balls are drawn one after the other without
replacement. Find the probability that both balls drawn are red. Enter the answer correct to two
decimal places.
Published solution
**1. First red draw:** Five of the twelve balls are red, so P(first red)=5/12.
**2. Update after success:** Without replacement, four red balls remain among eleven total balls. Thus P(second red given first red)=4/11.
**3. Multiply conditional probabilities:**
\[P(\text{both red})=\frac5{12}\times\frac4{11}=\frac5{33}=0.151515\ldots.\]
Rounded to two decimal places, this is 0.15. Squaring 5/12 would incorrectly treat the draws as if replacement occurred.
**Answer: 0.15.**
An escape room contains 7 unique keys. How many different 4-key sequences can be formed
using these keys without repetition?
Published solution
**1. Recognize ordered selection:** A four-key sequence specifies the first, second, third and fourth keys, so different orders count separately.
**2. Decrease the available choices:** Without repetition the positions have 7, 6, 5 and 4 choices respectively.
**3. Multiply:** \(7\times6\times5\times4=840\), also written \({}_7P_4=7!/(7-4)!\). Choosing four keys with a combination alone would omit their order.
**Answer: 840.**
A library contains 7 different Mathematics books and 6 different Statistics books. A student selects
3 Mathematics books and 2 Statistics books. The selected books are then arranged on a shelf such
that no two Statistics books are placed next to each other. Find the number of ways in which the
selected books can be arranged under the given condition.
Published solution
**1. Select the books:** Choose three Mathematics books in \(\binom73=35\) ways and two Statistics books in \(\binom62=15\) ways.
**2. Arrange the Mathematics books:** The three chosen books have 3!=6 shelf orders. Each order creates four gaps: before, between and after the Mathematics books.
**3. Place the Statistics books:** To keep them nonadjacent, use two different gaps. Choose those gaps in \(\binom42=6\) ways and assign the two distinct Statistics books in 2!=2 orders.
**4. Multiply all independent choices:** \(35\times15\times6\times6\times2=37,800\).
**Answer: 37800.**
A 6-digit security code is formed using digits from the set {0, 1, 2, 3, 4, 5, 6, 7, 8}, without
repetition. Determine the number of valid security codes in which:
-- the first digit is not zero,
-- the digits are either in strictly increasing order or strictly decreasing order, and
-- the digit 8 is included in the code.
Published solution
**1. Increasing codes:** A strictly increasing code is determined by its set of six digits. If zero were present it would be first, which is forbidden. Include 8 and select the other five digits from 1 through 7: \(\binom75=21\).
**2. Decreasing codes:** Include 8 and choose the other five digits from 0 through 7: \(\binom85=56\). Zero is allowed here because it appears last; 8 is first.
**3. Add disjoint cases:** A six-digit sequence of distinct digits cannot be both increasing and decreasing. Therefore 21+56=77 codes are valid. Do not multiply by 6!: each selected set has exactly one order of each specified type.
**Answer: 77.**
At a school, two teachers share the responsibility of creating quiz papers for students. Mr.
Chandra creates 35% of all quiz papers, while Ms. Sharma creates the remaining 65% of all quiz
papers. The passing rates differ between the two teachers, with only 20% of students passing
quizzes created by Mr. Chandra, whereas 80% of students pass quizzes created by Ms. Sharma.
Given that the student passed the quiz, what is the probability that the quiz was prepared by Mr.
Chandra? Enter the answer correct to two decimal places.
Published solution
**1. Define events:** Let C mean the paper was prepared by Mr. Chandra and P mean the student passed. Given P(C)=0.35, P(P|C)=0.20, P(Sharma)=0.65 and P(P|Sharma)=0.80.
**2. Find the overall pass probability:** \(P(P)=0.35(0.20)+0.65(0.80)=0.07+0.52=0.59\).
**3. Apply Bayes' rule:**
\[P(C\mid P)=\frac{P(C)P(P\mid C)}{P(P)}=\frac{0.07}{0.59}=\frac7{59}\approx0.118644.\]
To two decimal places, enter 0.12. The conditional passing rate 0.20 is not the requested reverse conditional probability.
**Answer: 0.12.**
**1. Count each event:** A standard deck has four Kings and thirteen Hearts.
**2. Correct the overlap:** The King of Hearts belongs to both groups and must be counted only once. Thus the union has 4+13-1=16 cards.
**3. Divide by deck size:** \(P(K\cup H)=16/52=4/13\). The word or means inclusive union, so a card satisfying both descriptions is allowed.
**Answer: B.**
A: **Incorrect:** 17/52 double-counts the King of Hearts.
B: **Correct:** The overlap correction gives 16 favorable cards out of 52.
C: **Incorrect:** 1/52 counts only the overlap, not the union.
D: **Incorrect:** 12/52 does not include all Kings and Hearts in the union.
A fair six-sided die is rolled once. Let A be the event that the outcome is even, and B be the event
that the outcome is strictly greater than 3. Which of the following statements are correct?
**1. List the events:** A={2,4,6} and B={4,5,6}. Both have probability 3/6=1/2.
**2. Compute overlap and union:** A intersection B={4,6}, so its probability is 2/6=1/3. A union B={2,4,5,6}, so its probability is 4/6=2/3.
**3. Distinguish a sum from a union:** P(A)+P(B)=1, but P(A union B) is smaller because the two overlapping outcomes must not be counted twice.
**Answer: A, C, D.**
A: **Correct:** A contains three of the six equally likely outcomes, giving 1/2.
B: **Incorrect:** The union has four outcomes, so it is 2/3 rather than 5/6.
C: **Correct:** The common outcomes 4 and 6 give probability 1/3.
D: **Correct:** Each marginal is 1/2, so their numerical sum is 1.
A movie streaming platform records that the probability a user watches a comedy movie is 0.45,
the probability the user watches an action movie is 0.50, and the probability that the user watches
neither a comedy movie nor an action movie is 0.18. Find the probability that the user watches
both a comedy movie and an action movie. Enter the answer correct to two decimal places.
Published solution
**1. Use the complement:** Watching neither genre is the complement of watching at least one, so P(C union A)=1-0.18=0.82.
**2. Apply inclusion-exclusion:** \(P(C\cup A)=P(C)+P(A)-P(C\cap A)\).
**3. Rearrange and substitute:** \(P(C\cap A)=0.45+0.50-0.82=0.13\). No independence assumption is given or needed; multiplying the two marginal probabilities would not be justified.
**Answer: 0.13.**
