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1. Define Random experiment ?
      Random experiment means all the outcomes of the experiment are known in advance but any specific outcome of the experiment is not known in advance.
      Eg: Tossing a coin is a random experiment because there are only two possible outcomes head and tail , and these outcomes are known well in advance. But the specific outcome of the experiment.ie, whether a head or a tail is not known in advance.

2. Define sample space?
      The set of all possible outcomes of an experiment is called sample space.

3. Write the sample space of the random experiment, when two coins are tossed together.
      When two coins are tossed together, the random experiment may result
      a) head (H) on the first coin and head (H) on the second coin.
      b) head (H) on the first coin and tail (T) on the second coin.
      c) tail (T) on the first coin and head (H) on the second coin.
      d) tail (T) on the first coin and tail (T) on the second coin.
      Thus the corresponding sample, S = {(H,H), (H,T), (T,H), (T,T) }

4. If two dice are rolled together then write the sample space for the random experiment?
      Since two dice are rolled together , the number of elements in the sample space is 36.
      S = {(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) }

5. Write the probability for the following question. If a dice is rolled once, an even number is required on the upper face of it.
      Total number of outcomes = 6
      Favourable outcomes = 2, 4, 6
      Number of favourable outcomes = 3
      ∴ Probability of getting an even number on the upper face.

     

6. Two dice are thrown simultaneously. Find the probability of getting four as the product?
       Let S be the sample space.Then n (S) = 36.
       Let A be the event getting four as product ie, {(1,4), (2,2), (4,1)}
       ∴ n (A) = 3

     

7. One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn. Find the probability that : a) The card drawn is black        b) The card drawn is a king.
      a) A pack of 52 cards has 26 black cards. So one black card may be drawn in 26 ways.
           ∴ Probability of drawing a black card = 26/52 = 1/2.
      b) A pack of 52 cards has 4 kings. So one king card may be drawn in 4 ways.
          ∴ Probability of drawing a king = 4/52 = 1/13.

8. One card is drawn from a pack of 52 cards ,each of the 52 cards  being equally likely to be drawn. Find the probability that the card drawn is black and a king?
     A pack of 52 cards has 2 black king cards. So one black king card may be drawn in 2 ways.
     ∴ Probability of drawing a black king card = 2/52.

9. A coin is tossed twice. What is the probability that at least one tail occurs?
        Here S = {HH, HT, TH, TT}
        ∴ number of possible outcomes n(S) = 4
        Let E be the event of getting at least one tail
        ∴ n(E) = 3
      ∴ Probability of getting at least one tail

10. A die is thrown, find the probability of following events.
      a) A prime number will appear.
      b) A number greater than or equal to 3 will appear.

        Here the sample space S = {1,2,3,4,5,6} ⇒ n (S) = 6
        a) Let A be the event of getting a prime number
            A = {2, 3, 5 } ⇒ n (A) = 3
           Thus
        b) Let B be the event of getting a number greater than or equal to 3
            B = {3, 4, 5, 6}  ⇒ n(B) = 4

            Thus

11. A die is thrown, find the probability of following events
      (i) A number less than or equal to one will appear.
      (ii) A number more than 6 will appear.
      (iii) A number less than 6 will appear.

              Here the sample space S = {1, 2, 3, 4, 5, 6} ⇒ n (S) = 6

              (i) Let C be the event of getting a number less than or equal to 1
                  C = { 1 } ⇒ n(C ) = 1

                 Thus

              (ii) Let D be the event of getting a number more than 6.
                   D = Φ ⇒ n (D) = 0

                    Thus

              (iii) Let X be the event of getting a number less than 6.
                   ∴ X = {1,2,3,4,5} ⇒ n (X) = 5
                  Thus

12. A card is selected from a pack of 52 cards, then how many points are there in the sample space ?
        One card is drawn from a pack of 52 cards ⇒ Number of points in the sample space S = n (S) = 52.

13. A card is selected from a pack of 52 cards.Calculate the probability that the card is an ace of spades ?
        Let A be the event of drawing an ace of spades. Now there is only one ace of spade. ∴ P(A) = 1/52.

14. Three coins are tossed once. Find the probability of getting (i) 3 heads    (ii) 3 tails
        Here three coins are tossed once.The sample space S is
        S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT } n (S) = 8
        (i) Let A be the event of getting 3 heads
            A = {HHH} ∴ n (A) = 1

            Thus

        (ii) Let T be the event of getting 3 tails
             T = {TTT} ⇒ n (T) = 1
              Thus P(T) = 1/8

15. A card is selected from a pack of 52 cards. Calculate the probability that the card is (i) an ace (ii)  black card
        (i) Let A be the event of drawing an ace. Now there are four aces
            ∴ P(B) = 4/52 = 1/13.
       (ii) Let B be the event of drawing a black card. Now there are 26 black cards.
            ∴ P(B) = 26/52 = 1/2.

16. Three coins are tossed once. Find the probability of getting a) no head    b) no tail.
         Here three coins are tossed once.The sample space S is
         S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT }
         ∴  n (S) = 8
         a) Let E be the event of getting no head
             E = TTT ⇒ n (E) = 1

             Thus

         b) Let F be the event of getting no tail
             F = HHH ⇒ n (F) = 1

            Thus

17. A letter is chosen at random from the word "ASSASSINATION". Find the probability that letter is  a) a vowel         b) a consonant.
       There are 13 letters in the word ASSASSINATION, which contains 6 vowels and 7 consonants.
       One letter is selected out of 13 letters in 13 C, = 13 ways
       a) One vowel is selected out of 6 vowels in 6 C, = 6 ways
           Thus probability of a vowel = 6/13
       b) One consonant is selected out of 7 consonants in 7 C, = 7 ways
           Thus probability of a consonant = 7/13.

18. Define equally likely events?
         Equally likely events are events that have the same theoretical probability (or likelihood) of occuring.

19. A coin is tossed twice. If the second toss results in a head, then a die is rolled. Write the sample space of the experiment?
    
A coin is tossed twice then S1 = {HH, HT, TH, TT}. It is given that when the second toss results a head, then a die is rolled.
       Only {HH,TH} show head in second toss. Hence S = {HT, TT, HH1, HH2, HH3, HH4, HH5, HH6, TH1, TH2, TH3, TH4, TH5, TH6 }.

20. From a group of 2 boys and 3 girls two children are selected. Find the sample space of the experiment ?
         Let two boys taken as B1, and B2 and three girls be G1,G2 and G3. Now two children can be chosen from given children. Hence S = {B1B2, B1G1, B1G2, B1G3, B2G1, B2G2, B2G3, G1G2, G1G3, G2G3 }

21. A box contains 1 red and 3 identical white balls. Two balls are drawn at random in succession without replacement. Write the sample space for this experiment?
        The four balls in the box are R,W,W,W.
        Two balls are drawn without replacement. Hence S = {RW, WR, WW}

22. In a single throw of a dice. Find the probability of getting a number greater than 2?
         In a single throw of a dice, the total possible outcomes are 6 (1,2,3,4,5 and 6 )
         Out of 1, 2, 3, 4, 5 and 6, the numbers greater than 2 are 3, 4, 5 and 6.
         Total number of favourable outcomes = 4
         ∴ P (getting a number greater than 2)

         

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