Probability Problems Based on Classical Definition of Probability

Data Science and A.I. Lecture Series

 

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Questions

• What is the total number of outcomes (sample space)?
• How do we determine favorable cases?
• How do probability rules apply to the problem?

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Example: Throwing Two Dice

Find the probability of:

1. A doublet (same number on both dice)
2. Sum equal to 7
3. Sum greater than 8
4. 3 on the first die and a multiple of 2 on the second die
5. Prime number on the first die and odd prime on the second die

Solution:

• \( P(A) = \frac{6}{36} = \frac{1}{6} \)
• \( P(B) = \frac{6}{36} = \frac{1}{6} \)
• \( P(C) = \frac{10}{36} = \frac{5}{18} \)
• \( P(D) = \frac{3}{36} = \frac{1}{12} \)
• \( P(E) = \frac{6}{36} = \frac{1}{6} \)

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Example: Drawing a Card

Find the probability of:

1. Drawing a red card
2. Drawing a face card
3. Drawing a spade card
4. Drawing a card other than clubs
5. Drawing a king

Solution:

• \( P(A) = \frac{26}{52} = \frac{1}{2} \)
• \( P(B) = \frac{12}{52} = \frac{3}{13} \)
• \( P(C) = \frac{13}{52} = \frac{1}{4} \)
• \( P(D) = \frac{39}{52} = \frac{3}{4} \)
• \( P(E) = \frac{4}{52} = \frac{1}{13} \)

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Example: Two Children

Find the probability of:

1. Elder child is a girl
2. Younger child is a girl
3. Both are girls
4. Both are of opposite sexes

Solution:

• \( P(A) = \frac{2}{4} = \frac{1}{2} \)
• \( P(B) = \frac{2}{4} = \frac{1}{2} \)
• \( P(C) = \frac{1}{4} \)
• \( P(D) = \frac{2}{4} = \frac{1}{2} \)

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Example: 53 Sundays in a Non-Leap Year

Find the probability of getting 53 Sundays.

Solution:

• Number of days in a non-leap year: 365
• Excess day options: {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}
• Favorable outcome: Sunday
• \( P(A) = \frac{1}{7} \)

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Example: Letters in “STATISTICS”

What is the probability of choosing a vowel?

Solution:

• Total letters: 10
• Vowels: A, I, I (3 vowels)
• \( P(A) = \frac{3}{10} \)

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Example: Horse Racing

Three horses A, B, and C are in a race. A is twice as likely to win as B, and B is twice as likely to win as C. Find their respective probabilities.

Solution:

• Let \( P(C) = p \)
• \( P(B) = 2p \), \( P(A) = 4p \)
• Using \( P(A) + P(B) + P(C) = 1 \):
• \( 4p + 2p + p = 1 \)
• Solve for \( p \): \( 7p = 1 \Rightarrow p = \frac{1}{7} \)
• Final probabilities:

\( P(A) = \frac{4}{7}, \ P(B) = \frac{2}{7}, \ P(C) = \frac{1}{7} \)

 

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Probability Examples

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