1. A box contains 10 red marbles, 20 blue marbles and 30 green marbles. 5 marbles are drawn from the box. What is the probability that at least one will be blue? 2. 4 cards are dra... 1. A box contains 10 red marbles, 20 blue marbles and 30 green marbles. 5 marbles are drawn from the box. What is the probability that at least one will be blue? 2. 4 cards are drawn from a well shuffled deck of 52 cards. What is the probability of obtaining: (i) all Kings, (ii) 3 Kings and 4 Queens, (iii) 1 King, 2 Queens and 3 Jacks, (iv) at least 3 Kings? 3. In a lot of 12 microwave ovens, there are 3 defective units. A person has ordered 4. If the selection is random, what is the probability that (i) all 4 units are good, (ii) at least 2 units are good? 4. A word consists of 9 letters, 5 consonants and 4 vowels. Three letters are chosen at random. What is the probability that more than one vowel will be selected? 5. A committee of two persons is selected from two men and two women. What is the probability that the committee will have (i) no man, (ii) one man, (iii) two men. Read more

Steps to Solve

Problem 1: Probability of At Least One Blue Marble

1. Calculate total marbles. There are 10 red, 20 blue, and 30 green marbles. $$ \text{Total} = 10 + 20 + 30 = 60 $$

2. Find the probability of no blue marbles. If 5 marbles are drawn without blue:

   • Total non-blue marbles = 40
   • Use the combination formula: $$ P(\text{no blue}) = \frac{\binom{40}{5}}{\binom{60}{5}} $$

3. Find the probability of at least one blue marble. $$ P(\text{at least one blue}) = 1 - P(\text{no blue}) $$

Problem 2: Probability with Playing Cards

1. Total ways to draw 4 cards. $$ \text{Total ways} = \binom{52}{4} $$

2. Probability calculations for all scenarios:

   • (i) All Kings: $$ P(\text{all Kings}) = \frac{\binom{4}{4}}{\binom{52}{4}} $$

   • (ii) 3 Kings and 1 Queen: $$ P(3K, 1Q) = \frac{\binom{4}{3} \cdot \binom{4}{1}}{\binom{52}{4}} $$

   • (iii) 1 King, 2 Queens, and 1 Jack: $$ P(1K, 2Q, 1J) = \frac{\binom{4}{1} \cdot \binom{4}{2} \cdot \binom{4}{1}}{\binom{52}{4}} $$

   • (iv) At least 3 Kings: $$ P(\geq 3K) = P(3K, 1X) + P(4K) $$

Problem 3: Microwave Ovens

1. Total selections of 4 ovens from 12: $$ \text{Total selections} = \binom{12}{4} $$

2. Calculating good and defective units:

   • (i) All 4 units are good: $$ P(\text{all good}) = \frac{\binom{9}{4}}{\binom{12}{4}} $$

   • (ii) At least 2 units are good: This requires calculating probabilities for 2, 3, and 4 good units.

Problem 4: Selecting Letters

1. Total ways to choose 3 letters. $$ \text{Total ways} = \binom{9}{3} $$

2. Calculate the scenarios for more than one vowel:

   • (i) 2 vowels, 1 consonant: $$ P(2V, 1C) = \frac{\binom{4}{2} \cdot \binom{5}{1}}{\binom{9}{3}} $$

   • (ii) 3 vowels: $$ P(3V) = \frac{\binom{4}{3}}{\binom{9}{3}} $$

   • Total probability: $$ P(\text{more than 1 vowel}) = P(2V, 1C) + P(3V) $$

Problem 5: Forming a Committee

1. Total ways to select 2 persons. $$ \text{Total ways} = \binom{4}{2} $$

2. Probability calculations for each scenario:

   • (i) No man: 2 women $$ P(\text{no man}) = \frac{\binom{2}{2}}{\binom{4}{2}} $$

   • (ii) One man: 1 man and 1 woman $$ P(1M) = \frac{\binom{2}{1} \cdot \binom{2}{1}}{\binom{4}{2}} $$

   • (iii) Two men: $$ P(2M) = \frac{\binom{2}{2}}{\binom{4}{2}} $$