Miscellaneous Exercise on Chapter 13 Solution
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Miscellaneous Exercise on Chapter 13 Solution

Question 1.

(i) Since A is a subset of B, A ∩ B = A. Therefore, P(B|A) = P(A) / P(A) = 1.

(ii) Since A and B are mutually exclusive, P(A ∩ B) = 0. Therefore, P(B|A) = 0 / P(A) = 0.

Question 2.

(i) Let A be the event of having a male child. P(A) = 3/4 since at least one child is male. The probability of both children being male is the probability of the second child being male given that the first child is male, which is P(A) = 3/4.

(ii) Let B be the event of having a female child. The probability of both children being female is the probability of the second child being female given that the first child is female, which is P(B) = 1/2.

Question 3.

Since there are an equal number of men and women, the probability of selecting a man is 1⁄2. Therefore, the probability of a grey-haired person being male is

0.05⁄0.05 + 0.0025 = 0.05⁄0.0525

Question 4.

Using the binomial probability formula, the probability of at most 6 right-handed people out of 10 is 

    P( X ≤ 6)   = 
  
    ∑ 
    
      x 
      = 
      0 
    
    6 
  
  
    
  
    
      (
      
        10 
        x 
      
      )
    
  


    × 
    
      
        0.9 
      
      
        x 
      
    
    × 
    
      
        0.1 
      
      
        10 
        - 
        x

Question 5.

The leap year can occur once every 4 years. Therefore, in 4 years, there are 3 non-leap years and 1 leap year. Each non-leap year has 52 weeks with 1 Tuesday each, totaling 156 Tuesdays. The leap year has an extra day, so it has 53 Tuesdays. Therefore, the probability is ( 1/4 ).

Question 6.

For each box, calculate the probability of drawing a red marble from that box, and then use Bayes' theorem to find the probability of drawing from each box given that the marble drawn is red.

Question 7.

Use the law of total probability to find the probability of suffering a heart attack given that the patient followed a course of meditation and yoga.

Question 8.

Enumerate all possible outcomes for the values of the determinant and count the number of positive outcomes to find the probability.

Question 9.

Use the conditional probability formula to find the probability of event A given event B.

Question 10.

Use Bayes' theorem to find the probability of the transferred ball being black given that the drawn ball from Bag II is red.

Question 11.

  • If ( P(A) ≠ 0 ) and ( P(B | A) = 1 ), it implies that event ( B ) always occurs given that event ( A ) has occurred. Therefore, ( B ) must be a superset of ( A ). Hence, the correct answer is option (B) ( B subset A ).

    • Question 12.

  • If ( P(A|B) > P(A) ), it means that the probability of event ( A ) occurring given that event ( B ) has occurred is greater than the probability of event ( A ) occurring independently. This implies that event ( B ) has some influence on event ( A ). Therefore, the correct answer is option (B) ( P(A | B) < P(A) * P(B) ).

    • Question 13.

  • If ( P(A) + P(B) - P(A | B) = P(A) ), it indicates that the probability of event ( A ) occurring and event ( B ) occurring minus the probability of event ( A ) occurring given that event ( B ) has occurred is equal to the probability of event ( A ) occurring. Simplifying, we find that ( P(B | A) = 1 ). Therefore, the correct answer is option (A) ( P(B|A) = 1 ).