Questions on Probability and Statistics
B.Tech Probability Questions with Detailed Solutions
Topic: Simple / Classical Probability
Level: B.Tech Engineering Mathematics
Marks: 6-10 Marks Each
Question 1: Probability Based on Playing Cards
A card is drawn at random from a well-shuffled standard deck of 52 playing cards. Find the probability that the card drawn is:
- a king or a queen,
- a red face card,
- neither an ace nor a king,
- either a heart or a face card.
Show Detailed Solution
Total number of cards in a standard deck:
\( n(S) = 52 \)
The classical definition of probability is:
\[ P(E)=\frac{\text{Number of favourable outcomes}} {\text{Total number of equally likely outcomes}} \]
(a) Probability of drawing a King or Queen
Number of kings = 4
Number of queens = 4
Therefore,
\[ n(E)=4+4=8 \]
Hence,
\[ P(\text{King or Queen})=\frac{8}{52} \]
\[ =\frac{2}{13} \]
Answer: \( \boxed{\frac{2}{13}} \)
(b) Probability of drawing a Red Face Card
Face cards are Jack, Queen and King.
There are 3 face cards in each suit.
The two red suits are Hearts and Diamonds.
Therefore, total red face cards:
\[ 3+3=6 \]
Hence,
\[ P(\text{Red Face Card})=\frac{6}{52} \]
\[ =\frac{3}{26} \]
Answer: \( \boxed{\frac{3}{26}} \)
(c) Probability of drawing neither an Ace nor a King
Number of aces = 4
Number of kings = 4
Therefore, total number of aces or kings:
\[ 4+4=8 \]
Number of cards which are neither ace nor king:
\[ 52-8=44 \]
Therefore,
\[ P(\text{Neither Ace nor King})=\frac{44}{52} \]
\[ =\frac{11}{13} \]
Answer: \( \boxed{\frac{11}{13}} \)
(d) Probability of drawing either a Heart or Face Card
Let:
\( A = \) event of drawing a heart
\( B = \) event of drawing a face card
Number of hearts:
\[ n(A)=13 \]
Number of face cards:
\[ n(B)=12 \]
There are 3 cards which are both hearts and face cards: Jack of Hearts, Queen of Hearts and King of Hearts.
Therefore,
\[ n(A\cap B)=3 \]
Using inclusion-exclusion:
\[ n(A\cup B)=n(A)+n(B)-n(A\cap B) \]
\[ =13+12-3=22 \]
Hence,
\[ P(A\cup B)=\frac{22}{52} \]
\[ =\frac{11}{26} \]
Answer: \( \boxed{\frac{11}{26}} \)
Question 2: Probability Based on Two Dice
Two fair dice are thrown simultaneously. Find the probability that:
- the sum of the numbers obtained is 8,
- the sum is greater than 9,
- the numbers on the two dice are equal,
- the product of the numbers obtained is even.
Show Detailed Solution
Each die has 6 possible outcomes.
Therefore, total number of outcomes:
\[ n(S)=6\times6=36 \]
(a) Sum is 8
The favourable outcomes are:
\[ (2,6),(3,5),(4,4),(5,3),(6,2) \]
Number of favourable outcomes = 5
Therefore,
\[ P(\text{Sum}=8)=\frac{5}{36} \]
Answer: \( \boxed{\frac{5}{36}} \)
(b) Sum greater than 9
Possible sums greater than 9 are 10, 11 and 12.
For sum 10:
\[ (4,6),(5,5),(6,4) \]
Number of outcomes = 3
For sum 11:
\[ (5,6),(6,5) \]
Number of outcomes = 2
For sum 12:
\[ (6,6) \]
Number of outcomes = 1
Total favourable outcomes:
\[ 3+2+1=6 \]
Therefore,
\[ P(\text{Sum}>9)=\frac{6}{36} \]
\[ =\frac{1}{6} \]
Answer: \( \boxed{\frac{1}{6}} \)
(c) Both dice show equal numbers
The favourable outcomes are:
\[ (1,1),(2,2),(3,3),(4,4),(5,5),(6,6) \]
Total favourable outcomes = 6
Therefore,
\[ P(\text{Equal Numbers})=\frac{6}{36} \]
\[ =\frac{1}{6} \]
Answer: \( \boxed{\frac{1}{6}} \)
(d) Product is even
The product will be odd only when both dice show odd numbers.
Odd numbers on a die are:
\[ 1,3,5 \]
Number of ways both dice show odd numbers:
\[ 3\times3=9 \]
Therefore,
\[ P(\text{Product Odd})=\frac{9}{36} \]
Hence,
\[ P(\text{Product Even}) =1-\frac{9}{36} \]
\[ =\frac{27}{36} =\frac{3}{4} \]
Answer: \( \boxed{\frac{3}{4}} \)
Question 3: Probability Based on Balls
A bag contains 7 red, 5 white and 4 blue balls. Three balls are drawn simultaneously at random. Find the probability that:
- all three balls are red,
- exactly two balls are red,
- one ball of each colour is drawn,
- at least one ball is blue.
Show Detailed Solution
Total number of balls:
\[ 7+5+4=16 \]
Number of ways to select 3 balls:
\[ {16 \choose 3} =\frac{16\times15\times14}{3\times2\times1} =560 \]
(a) All three balls are red
Number of ways to select 3 red balls from 7:
\[ {7 \choose 3}=35 \]
Therefore,
\[ P(\text{All Red})=\frac{35}{560} \]
\[ =\frac{1}{16} \]
Answer: \( \boxed{\frac{1}{16}} \)
(b) Exactly two balls are red
Choose 2 red balls from 7:
\[ {7 \choose 2}=21 \]
The third ball must be chosen from the 9 non-red balls.
\[ {9 \choose 1}=9 \]
Number of favourable selections:
\[ 21\times9=189 \]
Therefore,
\[ P(\text{Exactly 2 Red})=\frac{189}{560} \]
\[ =\frac{27}{80} \]
Answer: \( \boxed{\frac{27}{80}} \)
(c) One ball of each colour
Choose 1 red ball:
\[ {7 \choose 1}=7 \]
Choose 1 white ball:
\[ {5 \choose 1}=5 \]
Choose 1 blue ball:
\[ {4 \choose 1}=4 \]
Total favourable selections:
\[ 7\times5\times4=140 \]
Therefore,
\[ P(\text{One of Each Colour})=\frac{140}{560} \]
\[ =\frac{1}{4} \]
Answer: \( \boxed{\frac{1}{4}} \)
(d) At least one ball is blue
We use the complement rule:
\[ P(\text{At least one blue}) =1-P(\text{No blue}) \]
Number of non-blue balls:
\[ 7+5=12 \]
Number of ways to choose 3 non-blue balls:
\[ {12 \choose 3}=220 \]
Therefore,
\[ P(\text{At least one blue}) =1-\frac{220}{560} \]
\[ =\frac{340}{560} =\frac{17}{28} \]
Answer: \( \boxed{\frac{17}{28}} \)
Question 4: Probability Based on Five Coin Tosses
Five fair coins are tossed simultaneously. Find the probability of obtaining:
- exactly three heads,
- at least four heads,
- at most one head,
- at least one head and at least one tail.
Show Detailed Solution
Each coin has two possible outcomes: Head or Tail.
Therefore, total number of outcomes when 5 coins are tossed:
\[ n(S)=2^5=32 \]
(a) Exactly 3 heads
Number of ways to select the positions of 3 heads among 5 coins:
\[ {5 \choose 3}=10 \]
Therefore,
\[ P(\text{Exactly 3 Heads})=\frac{10}{32} \]
\[ =\frac{5}{16} \]
Answer: \( \boxed{\frac{5}{16}} \)
(b) At least 4 heads
At least 4 heads means either 4 heads or 5 heads.
Number of ways to get 4 heads:
\[ {5 \choose 4}=5 \]
Number of ways to get 5 heads:
\[ {5 \choose 5}=1 \]
Total favourable outcomes:
\[ 5+1=6 \]
Therefore,
\[ P(\text{At least 4 Heads})=\frac{6}{32} \]
\[ =\frac{3}{16} \]
Answer: \( \boxed{\frac{3}{16}} \)
(c) At most one head
At most one head means either 0 heads or 1 head.
Number of ways to get 0 heads:
\[ {5 \choose 0}=1 \]
Number of ways to get 1 head:
\[ {5 \choose 1}=5 \]
Total favourable outcomes:
\[ 1+5=6 \]
Hence,
\[ P(\text{At most 1 Head})=\frac{6}{32} \]
\[ =\frac{3}{16} \]
Answer: \( \boxed{\frac{3}{16}} \)
(d) At least one head and at least one tail
The only outcomes which do not contain both a head and a tail are:
HHHHH and TTTTT
Thus, number of unwanted outcomes = 2
Number of favourable outcomes:
\[ 32-2=30 \]
Therefore,
\[ P(\text{At least one H and one T})=\frac{30}{32} \]
\[ =\frac{15}{16} \]
Answer: \( \boxed{\frac{15}{16}} \)
Question 5: Probability Based on Selection of Students
A class consists of 8 boys and 6 girls. A committee of 4 students is selected at random. Find the probability that the committee contains:
- exactly 2 boys and 2 girls,
- all boys,
- at least one girl,
- more girls than boys.
Show Detailed Solution
Total number of students:
\[ 8+6=14 \]
Number of ways to choose a committee of 4 students:
\[ {14 \choose 4} =1001 \]
(a) Exactly 2 boys and 2 girls
Number of ways to choose 2 boys from 8:
\[ {8 \choose 2}=28 \]
Number of ways to choose 2 girls from 6:
\[ {6 \choose 2}=15 \]
Total favourable selections:
\[ 28\times15=420 \]
Therefore,
\[ P(\text{2 Boys and 2 Girls}) =\frac{420}{1001} \]
\[ =\frac{60}{143} \]
Answer: \( \boxed{\frac{60}{143}} \)
(b) All boys
Number of ways to choose 4 boys from 8:
\[ {8 \choose 4}=70 \]
Therefore,
\[ P(\text{All Boys})=\frac{70}{1001} \]
\[ =\frac{10}{143} \]
Answer: \( \boxed{\frac{10}{143}} \)
(c) At least one girl
Use the complement rule:
\[ P(\text{At least one girl}) =1-P(\text{No Girl}) \]
No girl means all 4 selected students are boys.
Therefore,
\[ P(\text{At least one girl}) =1-\frac{70}{1001} \]
\[ =\frac{931}{1001} \]
\[ =\frac{133}{143} \]
Answer: \( \boxed{\frac{133}{143}} \)
(d) More girls than boys
Since the committee contains 4 students, there are two possible cases:
Case 1: 3 girls and 1 boy
Case 2: 4 girls and 0 boys
Case 1:
Number of ways to select 3 girls:
\[ {6 \choose 3}=20 \]
Number of ways to select 1 boy:
\[ {8 \choose 1}=8 \]
Therefore:
\[ 20\times8=160 \]
Case 2:
Number of ways to select 4 girls:
\[ {6 \choose 4}=15 \]
Total favourable outcomes:
\[ 160+15=175 \]
Hence,
\[ P(\text{More Girls than Boys}) =\frac{175}{1001} \]
\[ =\frac{25}{143} \]
Answer: \( \boxed{\frac{25}{143}} \)
Important Formulae
Classical Probability:
\[ P(E)=\frac{n(E)}{n(S)} \]
Combination Formula:
\[ {n \choose r}=\frac{n!}{r!(n-r)!} \]
Complement Rule:
\[ P(A')=1-P(A) \]
Addition Rule:
\[ P(A\cup B)=P(A)+P(B)-P(A\cap B) \]
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