Questions on Probability and Statistics
Probability Problems with Detailed Solutions
Click each question to expand the detailed interpretation and solution.
Question 1: Two unbiased dice are thrown
Two fair dice are rolled together. Find the probability that:
(i) the sum is even
(ii) the sum is less than 5
(iii) the sum is at most 12
(i) the sum is even
(ii) the sum is less than 5
(iii) the sum is at most 12
Since each die has 6 possible outcomes, the total number of outcomes is:
$$6\times6=36$$
We calculate the number of favorable outcomes for each condition.
(i) Sum is even
The sum is even when:
Number of odd numbers on a die = 3 $$3\times3+3\times3=18$$ Therefore, $$P(\text{sum is even})=\frac{18}{36}=\frac12$$
(ii) Sum is less than 5
Possible sums are 2, 3 and 4. Number of favorable outcomes: $$1+2+3=6$$ Hence, $$P(\text{sum}<5)=\frac{6}{36}=\frac16$$
(iii) Sum is at most 12
The largest possible sum of two dice is 12, so all outcomes are favorable. $$P(\text{sum}\le 12)=1$$
The sum is even when:
- both numbers are even
- both numbers are odd
Number of odd numbers on a die = 3 $$3\times3+3\times3=18$$ Therefore, $$P(\text{sum is even})=\frac{18}{36}=\frac12$$
(ii) Sum is less than 5
Possible sums are 2, 3 and 4. Number of favorable outcomes: $$1+2+3=6$$ Hence, $$P(\text{sum}<5)=\frac{6}{36}=\frac16$$
(iii) Sum is at most 12
The largest possible sum of two dice is 12, so all outcomes are favorable. $$P(\text{sum}\le 12)=1$$
Final Answers:
\( P(\text{even})=\frac12 \)
\( P(<5)=\frac16 \)
\( P(\le12)=1 \)
\( P(\text{even})=\frac12 \)
\( P(<5)=\frac16 \)
\( P(\le12)=1 \)
Question 2: Four cards are drawn from a pack of 52 cards
Four cards are selected at random from a standard deck of 52 cards. Find the probability that:
(i) all four are aces
(ii) exactly three are kings and one is a queen
(iii) exactly three are black and one is red
(i) all four are aces
(ii) exactly three are kings and one is a queen
(iii) exactly three are black and one is red
The total number of ways of drawing 4 cards from 52 is:
$$^{52}C_4=270725$$
Each required event is counted separately and divided by total possible selections.
(i) All four are aces
Number of ways: $$^{4}C_4=1$$ Therefore, $$P=\frac{1}{270725}$$
(ii) Exactly three kings and one queen
Ways to choose 3 kings: $$^{4}C_3=4$$ Ways to choose 1 queen: $$^{4}C_1=4$$ Total favorable ways: $$4\times4=16$$ Thus, $$P=\frac{16}{270725}$$
(iii) Exactly three black and one red
Black cards = 26, Red cards = 26 $$^{26}C_3\times^{26}C_1$$ Hence, $$P=\frac{^{26}C_3\times^{26}C_1}{^{52}C_4}=\frac{104}{417}$$
Number of ways: $$^{4}C_4=1$$ Therefore, $$P=\frac{1}{270725}$$
(ii) Exactly three kings and one queen
Ways to choose 3 kings: $$^{4}C_3=4$$ Ways to choose 1 queen: $$^{4}C_1=4$$ Total favorable ways: $$4\times4=16$$ Thus, $$P=\frac{16}{270725}$$
(iii) Exactly three black and one red
Black cards = 26, Red cards = 26 $$^{26}C_3\times^{26}C_1$$ Hence, $$P=\frac{^{26}C_3\times^{26}C_1}{^{52}C_4}=\frac{104}{417}$$
Final Answers:
\( \frac1{270725} \)
\( \frac{16}{270725} \)
\( \frac{104}{417} \)
\( \frac1{270725} \)
\( \frac{16}{270725} \)
\( \frac{104}{417} \)
Question 3: Exactly 8 cards of the same suit in a bridge hand
In a bridge hand of 13 cards, find the probability of getting exactly 8 cards from one suit.
A bridge hand consists of 13 cards drawn from 52 cards.
We first count all possible bridge hands, then count the hands having exactly 8 cards from one suit.
Total number of bridge hands:
$$^{52}C_{13}$$
Choose one suit from 4:
$$^{4}C_1=4$$
Choose 8 cards from the 13 cards in that suit:
$$^{13}C_8$$
Choose remaining 5 cards from remaining 39 cards:
$$^{39}C_5$$
Thus,
$$P=\frac{4\binom{13}{8}\binom{39}{5}}{\binom{52}{13}}$$
Final Answer:
\( \frac{4\binom{13}{8}\binom{39}{5}}{\binom{52}{13}} \)
\( \frac{4\binom{13}{8}\binom{39}{5}}{\binom{52}{13}} \)
Question 4: At least one additional heart
A man is dealt 3 hearts from a standard pack of 52 cards. He is then given 4 more cards.
Find the probability that at least one of these additional 4 cards is also a heart.
Since the player already has 3 hearts, there are only 10 hearts left in the remaining 49 cards.
Instead of directly finding the probability of getting at least one heart, it is easier to find
the probability of getting no hearts in the next 4 cards, and subtract from 1.
Remaining cards = 49
Remaining hearts = 10
Remaining non-hearts = 39
Probability of no hearts in 4 additional cards: $$P(\text{no heart})=\frac{{^{39}C_4}}{{^{49}C_4}}$$ Therefore, $$P(\text{at least one heart})=1-\frac{{^{39}C_4}}{{^{49}C_4}}$$ $$P=1-\frac{82251}{211876}=0.6117$$
Remaining hearts = 10
Remaining non-hearts = 39
Probability of no hearts in 4 additional cards: $$P(\text{no heart})=\frac{{^{39}C_4}}{{^{49}C_4}}$$ Therefore, $$P(\text{at least one heart})=1-\frac{{^{39}C_4}}{{^{49}C_4}}$$ $$P=1-\frac{82251}{211876}=0.6117$$
Final Answer:
\( P = 0.6117 \)
\( P = 0.6117 \)
Question 5: Vowels together in "BALLOON"
If the letters of the word BALLOON are arranged at random,
what is the probability that all the vowels come together?
The word BALLOON has 7 letters with repeated letters:
L appears twice, and O appears twice.
We first find the total number of arrangements of the word. Then we treat the vowels AOO as one block and count the favorable arrangements.
L appears twice, and O appears twice.
We first find the total number of arrangements of the word. Then we treat the vowels AOO as one block and count the favorable arrangements.
Total arrangements of BALLOON:
$$\frac{7!}{2!2!}=1260$$
Vowels are A, O, O.
Treat them as one block:
$$(AOO), B, L, L, N$$
So we have 5 objects, with L repeated twice.
Number of arrangements:
$$\frac{5!}{2!}=60$$
The vowels AOO can be arranged among themselves in:
$$\frac{3!}{2!}=3$$
Therefore favorable arrangements:
$$60\times3=180$$
Hence,
$$P=\frac{180}{1260}=\frac17$$
Final Answer:
\( P = \frac17 \)
\( P = \frac17 \)
Question 6: Word arrangement problems
(a) If the letters of the word EDUCATION are arranged at random, find the probability that there are exactly 3 letters between E and N.
(b) If the letters of the word PARALLEL are arranged at random, find the probability that all three L’s come together.
(b) If the letters of the word PARALLEL are arranged at random, find the probability that all three L’s come together.
In both parts, we calculate:
Probability = Favorable arrangements / Total arrangements
For (a), we place E and N exactly 4 positions apart.
For (b), we treat the three L’s as one single block.
Probability = Favorable arrangements / Total arrangements
For (a), we place E and N exactly 4 positions apart.
For (b), we treat the three L’s as one single block.
(a) Word EDUCATION
EDUCATION has 9 distinct letters. Total arrangements: $$9!$$ To have exactly 3 letters between E and N, the pair (E,N) must occupy: $$(1,5),(2,6),(3,7),(4,8),(5,9)$$ So number of position choices = 5 E and N can interchange in: $$2!$$ Remaining 7 letters can be arranged in: $$7!$$ Favorable arrangements: $$5\times2\times7!$$ Hence, $$P=\frac{5\times2\times7!}{9!}=\frac{5}{36}$$
(b) Word PARALLEL
Letters: P, A, R, A, L, L, E, L Total arrangements: $$\frac{8!}{2!3!}$$ Treat the three L’s as one block: $$(LLL), P, A, R, A, E$$ This gives 6 objects with A repeated twice. Favorable arrangements: $$\frac{6!}{2!}$$ Therefore, $$P=\frac{\frac{6!}{2!}}{\frac{8!}{2!3!}}=\frac{3}{28}$$
EDUCATION has 9 distinct letters. Total arrangements: $$9!$$ To have exactly 3 letters between E and N, the pair (E,N) must occupy: $$(1,5),(2,6),(3,7),(4,8),(5,9)$$ So number of position choices = 5 E and N can interchange in: $$2!$$ Remaining 7 letters can be arranged in: $$7!$$ Favorable arrangements: $$5\times2\times7!$$ Hence, $$P=\frac{5\times2\times7!}{9!}=\frac{5}{36}$$
(b) Word PARALLEL
Letters: P, A, R, A, L, L, E, L Total arrangements: $$\frac{8!}{2!3!}$$ Treat the three L’s as one block: $$(LLL), P, A, R, A, E$$ This gives 6 objects with A repeated twice. Favorable arrangements: $$\frac{6!}{2!}$$ Therefore, $$P=\frac{\frac{6!}{2!}}{\frac{8!}{2!3!}}=\frac{3}{28}$$
Final Answers:
(a) \( \frac{5}{36} \)
(b) \( \frac{3}{28} \)
(a) \( \frac{5}{36} \)
(b) \( \frac{3}{28} \)
Question 7: Particular pair of books on a shelf
Ten books are placed at random on a shelf. Find the probability that a particular pair of books shall be:
(i) Always together
(ii) Never together
(i) Always together
(ii) Never together
We first count the total number of ways of arranging 10 books on a shelf:
$$10!$$
For part (i), we treat the particular pair as one unit.
For part (ii), we subtract the probability of being together from 1.
(i) Probability that the pair is always together
Treat the pair as one single block. Then the total objects are:
8 other books + 1 block = 9 objects Number of arrangements of these 9 objects: $$9!$$ The two books inside the pair can be arranged in: $$2!$$ Therefore favorable arrangements: $$9!\times2!$$ Probability: $$P=\frac{9!\times2!}{10!}=\frac{2}{10}=\frac15$$
(ii) Probability that the pair is never together
$$P(\text{never together})=1-\frac15=\frac45$$
Treat the pair as one single block. Then the total objects are:
8 other books + 1 block = 9 objects Number of arrangements of these 9 objects: $$9!$$ The two books inside the pair can be arranged in: $$2!$$ Therefore favorable arrangements: $$9!\times2!$$ Probability: $$P=\frac{9!\times2!}{10!}=\frac{2}{10}=\frac15$$
(ii) Probability that the pair is never together
$$P(\text{never together})=1-\frac15=\frac45$$
Final Answers:
(i) \( \frac15 \)
(ii) \( \frac45 \)
(i) \( \frac15 \)
(ii) \( \frac45 \)
Question 8: Four-digit number formed from 1,2,3,4,5
A four-digit number is formed using the digits 1, 2, 3, 4, 5 without repetition.
Find the probability that the number is even.
A number is even if its last digit is even.
Among the digits {1,2,3,4,5}, only 2 and 4 are even.
We first find the total number of 4-digit numbers, then count how many end with an even digit.
Total 4-digit numbers formed from 5 digits:
$$^5P_4=5\times4\times3\times2=120$$
To form an even number, the last digit must be 2 or 4.
So there are 2 choices for the last digit.
The remaining 3 places can be filled by the remaining 4 digits:
$$^4P_3=4\times3\times2=24$$
Favorable numbers:
$$2\times24=48$$
Therefore,
$$P=\frac{48}{120}=\frac25$$
Final Answer:
\( \frac25 \)
\( \frac25 \)
Question 9: Comparing chances of throwing 6, 7 and 10
Compare the chances of:
(i) throwing a 6 with one die
(ii) throwing a 7 with two dice
(iii) throwing a 10 with three dice
(i) throwing a 6 with one die
(ii) throwing a 7 with two dice
(iii) throwing a 10 with three dice
We calculate the probability of each event:
- Getting 6 on one die
- Getting sum 7 on two dice
- Getting sum 10 on three dice
- Getting 6 on one die
- Getting sum 7 on two dice
- Getting sum 10 on three dice
(i) Throwing a 6 with one die
$$P=\frac16$$
(ii) Throwing a sum of 7 with two dice Total outcomes: $$6\times6=36$$ Favorable outcomes: $$(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$$ Number of favorable outcomes = 6 $$P=\frac{6}{36}=\frac16$$
(iii) Throwing a sum of 10 with three dice Total outcomes: $$6^3=216$$ Number of favorable outcomes for sum 10 = 27 $$P=\frac{27}{216}=\frac18$$
(ii) Throwing a sum of 7 with two dice Total outcomes: $$6\times6=36$$ Favorable outcomes: $$(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$$ Number of favorable outcomes = 6 $$P=\frac{6}{36}=\frac16$$
(iii) Throwing a sum of 10 with three dice Total outcomes: $$6^3=216$$ Number of favorable outcomes for sum 10 = 27 $$P=\frac{27}{216}=\frac18$$
Final Answers:
(i) \( \frac16 \)
(ii) \( \frac16 \)
(iii) \( \frac18 \)
(i) \( \frac16 \)
(ii) \( \frac16 \)
(iii) \( \frac18 \)
Question 10: Product of two non-negative numbers
The sum of two non-negative numbers is \(S\).
Find the probability that their product is at least half of the maximum possible product.
Let the numbers be \(x\) and \(S-x\).
Their product is:
$$x(S-x)$$
The maximum product occurs when the numbers are equal:
$$x=\frac{S}{2}$$
Maximum product:
$$\frac{S^2}{4}$$
We want the probability that:
$$x(S-x)\ge\frac12\times\frac{S^2}{4}=\frac{S^2}{8}$$
Solve:
$$x(S-x)\ge\frac{S^2}{8}$$
$$Sx-x^2\ge\frac{S^2}{8}$$
Rearranging:
$$x^2-Sx+\frac{S^2}{8}\le0$$
Solving gives:
$$x\in\left[\frac{S}{2}\left(1-\frac{1}{\sqrt2}\right),\frac{S}{2}\left(1+\frac{1}{\sqrt2}\right)\right]$$
Length of favorable interval:
$$\frac{S}{\sqrt2}$$
Total interval length:
$$S$$
Therefore,
$$P=\frac{S/\sqrt2}{S}=\frac{1}{\sqrt2}$$
Final Answer:
\( \frac{1}{\sqrt2} \)
\( \frac{1}{\sqrt2} \)
Question 11: Sum of two dice
If two dice are thrown, find the probability that the sum is:
(i) less than 5
(ii) a multiple of 3
(i) less than 5
(ii) a multiple of 3
When two dice are thrown, total possible outcomes:
$$6\times6=36$$
We count favorable outcomes for each event.
(i) Sum less than 5
Possible sums: 2, 3, 4
Number of outcomes:
$$1+2+3=6$$
Therefore,
$$P=\frac{6}{36}=\frac16$$
(ii) Sum is a multiple of 3 Multiples of 3 are 3, 6, 9, 12 Number of outcomes: $$2+5+4+1=12$$ Therefore, $$P=\frac{12}{36}=\frac13$$
(ii) Sum is a multiple of 3 Multiples of 3 are 3, 6, 9, 12 Number of outcomes: $$2+5+4+1=12$$ Therefore, $$P=\frac{12}{36}=\frac13$$
Final Answers:
(i) \( \frac16 \)
(ii) \( \frac13 \)
(i) \( \frac16 \)
(ii) \( \frac13 \)
Question 12: Heart or Face Card
A card is drawn from a well-shuffled pack of 52 cards.
Find the probability that it is either a Heart or a Face card (King, Queen, Jack).
We use:
$$P(H\cup F)=P(H)+P(F)-P(H\cap F)$$
Hearts = 13 cards, Face cards = 12 cards, Heart face cards = 3 cards.
Favorable cards:
$$13+12-3=22$$
Therefore:
$$P=\frac{22}{52}=\frac{11}{26}$$
Final Answer:
\( \frac{11}{26} \)
\( \frac{11}{26} \)
Question 13: Marbles in a bag
A bag contains 5 red and 3 blue marbles. If 3 marbles are drawn at random,
find the probability that at least one is blue.
Instead of directly finding "at least one blue", we use complement:
$$P(\text{at least one blue})=1-P(\text{all red})$$
Total ways:
$$^8C_3$$
Ways to draw all red:
$$^5C_3$$
Therefore:
$$P=1-\frac{^5C_3}{^8C_3}$$
$$P=1-\frac{10}{56}=\frac{23}{28}$$
Final Answer:
\( \frac{23}{28} \)
\( \frac{23}{28} \)
Question 14: Marksmen hitting a target
Three marksmen take one shot each at a target.
Their probabilities of hitting are 0.6, 0.5, and 0.8.
Find the probability that the target is hit at least once.
It is easier to find the probability that all miss,
then subtract from 1.
Probability all miss:
$$(1-0.6)(1-0.5)(1-0.8)$$
$$=0.4\times0.5\times0.2=0.04$$
Therefore:
$$P(\text{at least one hit})=1-0.04=0.96$$
Final Answer:
\( 0.96 \)
\( 0.96 \)
Question 15: Showing \(P(A\cap B)\ge 0.5\)
If \(P(A)=0.7\) and \(P(B)=0.8\), show that:
$$P(A\cap B)\ge0.5$$
Use:
$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$
Since \(P(A\cup B)\le1\), we get a lower bound for \(P(A\cap B)\).
$$0.7+0.8-P(A\cap B)\le1$$
$$1.5-P(A\cap B)\le1$$
$$0.5\le P(A\cap B)$$
Hence:
$$P(A\cap B)\ge0.5$$
Final Answer:
\( P(A\cap B)\ge0.5 \)
\( P(A\cap B)\ge0.5 \)
Question 16: Last fused bulb
A box contains 10 bulbs, 3 fused.
Bulbs are tested one by one without replacement.
Find the probability that the 6th bulb tested is the last fused one.
For the 6th bulb to be the last fused:
- exactly 2 fused among first 5 bulbs
- 6th bulb is fused
Probability:
$$\frac{^3C_2\cdot ^7C_3}{^10C_6}$$
$$=\frac{3\times35}{210}=\frac12$$
Final Answer:
\( \frac12 \)
\( \frac12 \)
Question 17: First death among n men
\(p\) is the probability that a man aged \(x\) dies in a year.
Find the probability that among \(n\) men, \(A_1\) dies in a year and is the first to die.
If exactly \(r\) men die, then:
- \(A_1\) must be one of them
- \(A_1\) must be the first among the \(r\) deaths
The required probability is:
$$\frac1n\left[1-(1-p)^n\right]$$
Final Answer:
\( \frac1n\left[1-(1-p)^n\right] \)
\( \frac1n\left[1-(1-p)^n\right] \)
Question 18: Contradicting statements
Odds that A speaks truth are 4:1 and odds that B speaks truth are 3:2.
Find the percentage of cases in which they contradict each other.
They contradict when:
- A tells truth, B lies
- A lies, B tells truth
$$P(A_T)=\frac45,\quad P(B_T)=\frac35$$
Contradiction:
$$\frac45\cdot\frac25+\frac15\cdot\frac35$$
$$=\frac{8}{25}+\frac{3}{25}=\frac{11}{25}=44\%$$
Final Answer:
\(44\%\)
\(44\%\)
Question 19: Sustainable building design
Architects A, B, and C are selected in ratio 3:2:5.
Their probabilities of sustainable design are 0.4, 0.6, and 0.9.
Find the probability that the final project is sustainable.
Use weighted probability:
$$P(S)=P(A)P(S|A)+P(B)P(S|B)+P(C)P(S|C)$$
$$P(A)=\frac3{10},\quad P(B)=\frac2{10},\quad P(C)=\frac5{10}$$
$$P(S)=\frac3{10}(0.4)+\frac2{10}(0.6)+\frac5{10}(0.9)$$
$$=0.12+0.12+0.45=0.69$$
Final Answer:
\(0.69\)
\(0.69\)
Question 20: Screen defect probability
Lines X, Y, Z produce 50%, 30%, 20% of smartphones.
Their defect rates are 1%, 2%, 3%.
If a phone has a screen defect, find the probability it was manufactured by line X.
Use Bayes' theorem:
$$P(X|D)=\frac{P(X)P(D|X)}{P(D)}$$
$$P(D)=0.5(0.01)+0.3(0.02)+0.2(0.03)=0.017$$
$$P(X|D)=\frac{0.5\times0.01}{0.017}=\frac{5}{17}$$
Final Answer:
\( \frac{5}{17} \)
\( \frac{5}{17} \)
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