Math, asked by Kavyashree4468, 11 months ago

An unbiased coin is tossed. If the outcome is a head then a pair of unbiased dice is rolled and the sum of the numbers obtained on them is noted. If the toss of the coin results in tail then a card from a well-shuffled pack of nine cards numbered 1, 2, 3, ..., 9 is randomly picked and the number on the card is noted. The probability that the noted number is either 7 or 8 is:

Answers

Answered by prettystefina11
8

Answer:

1) An unbiased coin is tossed.

     Total number of possible outcomes = 2 (head,tail).

     The probability of getting a Head or a Tail is the same.

      The probability that the head is the outcome P(H) = 1/2

      The probability that the tail is the outcome P(T) = 1/2 [ Since P(H) + P(T) = 1]

2) If, the outcome is a head:      

   Given that; a pair of dice is rolled

   Total possible outcomes when a pair of dice is rolled = 36

    Which are ;

    {(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),

     (2,1),(2,2),(2,3),(2,4),(2,5),(2,6),

     (3,1),(3,2),(3,3),(3,4),(3,5),(3,6),

     (4,1),(4,2),(4,3),(4,4),(4,5),(4,6),

     (5,1),(5,2),(5,3),(5,4),(5,5),(5,6),

     (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

Probability of getting a sum of 7, P(S1) =        Number of occurrences with a sum of 7     /    Total number of occurrences                                               

                                                                  =  6/36

 Since, possible outcomes are ; {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)} = 6

Probability of getting a sum of 8, P(S2) =        Number of occurrences with a sum of 8     /    Total number of occurrences 

                                                                  =  5/36

 Since, possible outcomes are ; {(2,6),(3,5),(4,4),(5,3),(6,2)} = 5

So, the probability of getting a sum of 7 or 8  P(H1), when the unbiased coin tossed shows Head

                                           P(H1) = P(H).P(S1) + P(H).P(S2)

                                                     = 1/2 . 6/36 + 1/2  . 5/36

                                                     =    11/72

3) If, the outcome is a tail:


    Given that; a card is picked from a pack of cards 1,2,3,4,….9

   Total possible outcomes = 9

   {1,2,3,4,5,6,7,8,9}

   Probability of picking the number 7, P(C1) =  Number of   occurrences with the number 7  /  Total number of occurrences      

                                                                           =  1/9

 Probability of picking the number 8, P(C2) =  Number of   occurrences with the number 8  /  Total number of occurrences      

                                                                           =  1/9

So, the probability of getting the numbers 7 or 8 P(T1), when the unbiased coin tossed shows Tail

                                     P(T1)  = P(T).P(C1) + P(T).P(C2)

                                               = 1/2 . 1/9 + 1/2  . 1/9

                                               =   2/18

4) The probability that the required number noted is either 7 or 8 is:

  Since Point (2) and Point (3) are mutually exclusive events;

    The required probability is;

                                               P(H1) + P(T1)

                                            = 11/72 + 2/18 = (11+8)/72 = 19/72 =    0.264

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