Probability

Q1: A coin is tossed once. What is the probability of getting a head?

Step 1: Understanding the experiment
When a coin is tossed once, it has two possible outcomes.
These outcomes together form the sample space.
Sample Space, \( S = \{H, T\} \)
Where,
\( H \) = Head
\( T \) = Tail
Step 2: Finding favourable outcomes
The favourable outcome for getting a head is only \( H \).
Number of favourable outcomes \( = 1 \)
Total number of possible outcomes \( = 2 \)
Step 3: Applying probability formula
Probability of an event is given by:
\( P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \)
So,
\( P(\text{Head}) = \frac{1}{2} \)
Answer: \( P(\text{getting a head}) = \frac{1}{2} \)

Q2: A die is rolled once. What is the probability of getting:

i. an odd number?

Step 1: When a die is rolled once, the possible outcomes are the numbers from 1 to 6.
Sample Space, \( S = \{1, 2, 3, 4, 5, 6\} \)
Step 2: Odd numbers in the sample space are:
\( \{1, 3, 5\} \)
Number of favourable outcomes \( = 3 \)
Total number of possible outcomes \( = 6 \)
Step 3: Using probability formula:
\( P(\text{odd}) = \frac{3}{6} = \frac{1}{2} \)
Answer: \( P(\text{odd number}) = \frac{1}{2} \)

ii. a composite number?

Step 1: Composite numbers are numbers greater than 1 which are not prime.
Composite numbers in the sample space are:
\( \{4, 6\} \)
Number of favourable outcomes \( = 2 \)
Total number of possible outcomes \( = 6 \)
Step 2: Applying probability formula:
\( P(\text{composite}) = \frac{2}{6} = \frac{1}{3} \)
Answer: \( P(\text{composite number}) = \frac{1}{3} \)

iii. a number less than 5?

Step 1: Numbers less than 5 in the sample space are:
\( \{1, 2, 3, 4\} \)
Number of favourable outcomes \( = 4 \)
Total number of possible outcomes \( = 6 \)
Step 2: Calculating probability:
\( P(\text{number} \lt 5) = \frac{4}{6} = \frac{2}{3} \)
Answer: \( P(\text{number} \lt 5) = \frac{2}{3} \)

iv. a multiple of 3?

Step 1: Multiples of 3 in the sample space are:
\( \{3, 6\} \)
Number of favourable outcomes \( = 2 \)
Total number of possible outcomes \( = 6 \)
Step 2: Using probability formula:
\( P(\text{multiple of 3}) = \frac{2}{6} = \frac{1}{3} \)
Answer: \( P(\text{multiple of 3}) = \frac{1}{3} \)

Q3: Two coins are tossed simultaneously. Find the probability of getting:

i. exactly one head

Step 1: When two coins are tossed simultaneously, all possible outcomes are:
Sample Space, \( S = \{HH, HT, TH, TT\} \)
Total number of outcomes \( = 4 \)
Step 2: Exactly one head occurs in the outcomes:
\( \{HT, TH\} \)
Number of favourable outcomes \( = 2 \)
Step 3: Applying probability formula:
\( P(\text{exactly one head}) = \frac{2}{4} = \frac{1}{2} \)
Answer: \( P(\text{exactly one head}) = \frac{1}{2} \)

ii. at least one tail

Step 1: At least one tail means one or more tails.
Favourable outcomes are:
\( \{HT, TH, TT\} \)
Number of favourable outcomes \( = 3 \)
Total number of possible outcomes \( = 4 \)
Step 2: Calculating probability:
\( P(\text{at least one tail}) = \frac{3}{4} \)
Answer: \( P(\text{at least one tail}) = \frac{3}{4} \)

iii. no tail

Step 1: No tail means both outcomes are heads.
Favourable outcome is:
\( \{HH\} \)
Number of favourable outcomes \( = 1 \)
Total number of possible outcomes \( = 4 \)
Step 2: Applying probability formula:
\( P(\text{no tail}) = \frac{1}{4} \)
Answer: \( P(\text{no tail}) = \frac{1}{4} \)

iv. at most one head

Step 1: At most one head means zero head or one head.
Favourable outcomes are:
\( \{TT, HT, TH\} \)
Number of favourable outcomes \( = 3 \)
Total number of possible outcomes \( = 4 \)
Step 2: Calculating probability:
\( P(\text{at most one head}) = \frac{3}{4} \)
Answer: \( P(\text{at most one head}) = \frac{3}{4} \)

Q4: In a draw there are 10 prizes and 20 blanks. A ticket is chosen at random. What is the probability of getting a prize?

Step 1: Listing total outcomes
Total number of prize tickets \( = 10 \).
Total number of blank tickets \( = 20 \).
Total number of tickets in the draw is:
\( \text{Total tickets} = 10 + 20 = 30 \).
Step 2: Finding favourable outcomes
A favourable outcome means selecting a prize ticket.
Number of favourable outcomes \( = 10 \).
Step 3: Applying probability formula
Probability of an event is given by:
\( P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \).
Substituting the values,
\( P(\text{prize}) = \frac{10}{30} = \frac{1}{3} \).
Answer: \( P(\text{getting a prize}) = \frac{1}{3} \)

Q5: In a box of 100 electric bulbs, 8 bulbs are defective. If one bulb is taken out at random from the box, what is the probability that the bulb drawn is:

i. defective?

Step 1: Total number of electric bulbs in the box \( = 100 \).
Step 2: Number of defective bulbs \( = 8 \).
Step 3: Using the probability formula:
\( P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \)
Here,
Favourable outcome \( = \) defective bulb \( = 8 \).
Total outcomes \( = 100 \).
So,
\( P(\text{defective}) = \frac{8}{100} = \frac{2}{25} \)
Answer: \( P(\text{defective bulb}) = \frac{2}{25} \)

ii. non-defective?

Step 1: Number of non-defective bulbs is obtained by subtracting defective bulbs from total bulbs.
\( \text{Non-defective bulbs} = 100 – 8 = 92 \)
Step 2: Applying probability formula:
Favourable outcomes \( = 92 \).
Total outcomes \( = 100 \).
So,
\( P(\text{non-defective}) = \frac{92}{100} = \frac{23}{25} \)
Answer: \( P(\text{non-defective bulb}) = \frac{23}{25} \)

Q6: A letter is chosen at random from the letters of the word EXAMINATION. What is the probability that it is a vowel?

Step 1: Listing total outcomes
The given word is EXAMINATION.
Total number of letters in the word EXAMINATION are counted as follows:
E, X, A, M, I, N, A, T, I, O, N
Total number of letters \( = 11 \).
Step 2: Finding favourable outcomes
Vowels are the letters \( A, E, I, O, U \).
Vowels present in the word EXAMINATION are:
E, A, I, A, I, O
Number of vowels \( = 6 \).
Step 3: Applying probability formula
Probability of an event is given by:
\( P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \)
So,
\( P(\text{vowel}) = \frac{6}{11} \)
Answer: \( P(\text{getting a vowel}) = \frac{6}{11} \)

Q7: 19 cards, numbered 1, 2, 3, ……, 19 are put in a box and mixed thoroughly. One card is chosen at random from the box. What is the probability that the number on the chosen card is:

Step 1: The numbers written on the cards are from 1 to 19.
Sample Space,
\( S = \{1, 2, 3, 4, \ldots, 19\} \)
Total number of outcomes \( = 19 \).

i. a prime number?

Step 2: Prime numbers between 1 and 19 are:
\( \{2, 3, 5, 7, 11, 13, 17, 19\} \)
Number of favourable outcomes \( = 8 \).
Step 3: Applying probability formula:
\( P(\text{prime}) = \frac{8}{19} \)
Answer: \( P(\text{prime number}) = \frac{8}{19} \)

ii. a multiple of 4?

Step 1: Multiples of 4 between 1 and 19 are:
\( \{4, 8, 12, 16\} \)
Number of favourable outcomes \( = 4 \).
Total number of outcomes \( = 19 \).
Step 2: Calculating probability:
\( P(\text{multiple of 4}) = \frac{4}{19} \)
Answer: \( P(\text{multiple of 4}) = \frac{4}{19} \)

iii. a composite number less than 10?

Step 1: Composite numbers less than 10 are:
\( \{4, 6, 8, 9\} \)
Number of favourable outcomes \( = 4 \).
Total number of outcomes \( = 19 \).
Step 2: Applying probability formula:
\( P(\text{composite} \lt 10) = \frac{4}{19} \)
Answer: \( P(\text{composite number less than 10}) = \frac{4}{19} \)

iv. neither divisible by 3 nor 5?

Step 1: Numbers divisible by 3 are:
\( \{3, 6, 9, 12, 15, 18\} \)
Numbers divisible by 5 are:
\( \{5, 10, 15\} \)
Numbers divisible by 3 or 5 are:
\( \{3, 5, 6, 9, 10, 12, 15, 18\} \)
Step 2: Numbers neither divisible by 3 nor 5 are:
\( \{1, 2, 4, 7, 8, 11, 13, 14, 16, 17, 19\} \)
Number of favourable outcomes \( = 11 \).
Total number of outcomes \( = 19 \).
Step 3: Calculating probability:
\( P(\text{neither divisible by 3 nor 5}) = \frac{11}{19} \)
Answer: \( P(\text{neither divisible by 3 nor 5}) = \frac{11}{19} \)

Q8: A card is drawn from a well-shuffled pack of 52 cards. Find the probability that the card drawn is:

Step 1: A standard deck of playing cards contains 52 cards.
Total number of possible outcomes \( = 52 \).

i. a black king?

Step 2: There are two black suits: Spades and Clubs.
Black kings are:
King of Spades, King of Clubs
Number of favourable outcomes \( = 2 \).
Step 3: Applying probability formula:
\( P(\text{black king}) = \frac{2}{52} = \frac{1}{26} \)
Answer: \( P(\text{black king}) = \frac{1}{26} \)

ii. a jack?

Step 1: There are four jacks in a pack of cards.
Jack of Hearts, Jack of Diamonds, Jack of Clubs, Jack of Spades
Number of favourable outcomes \( = 4 \).
Total outcomes \( = 52 \).
Step 2: Calculating probability:
\( P(\text{jack}) = \frac{4}{52} = \frac{1}{13} \)
Answer: \( P(\text{jack}) = \frac{1}{13} \)

iii. a spade?

Step 1: Each suit contains 13 cards.
Number of spade cards \( = 13 \).
Total outcomes \( = 52 \).
Step 2: Applying probability formula:
\( P(\text{spade}) = \frac{13}{52} = \frac{1}{4} \)
Answer: \( P(\text{spade}) = \frac{1}{4} \)

iv. a king or a queen?

Step 1: Total number of kings in the deck \( = 4 \).
Total number of queens in the deck \( = 4 \).
Total favourable outcomes \( = 4 + 4 = 8 \).
Step 2: Applying probability formula:
\( P(\text{king or queen}) = \frac{8}{52} = \frac{2}{13} \)
Answer: \( P(\text{king or queen}) = \frac{2}{13} \)