Math, asked by Anonymous, 1 year ago

Chapter- Probability [ICSE CLASS X]
3,4,5 question.

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Anonymous: I will solve the problems a bit later

Answers

Answered by Shubhendu8898
6

3(a) Given,

         Probability of winning Ram is 0.48

(i)    Probability of not winning ram = (1-probability of winning Ram)

         = 1-0.48

Probability of winning Ram = 0.52

(ii)        Since there are only two participants.

   Therefore,

   Probability of winning Shyam = probability of not winning Ram =0.52

3(b) Given,

 No. of total bulbs = 600

Defective bulbs = 12

Non-Defective bulbs = 600-12

                                       =588

So,

 Probability of taking out non- defective bulb= No. of non-defective bulbs/total bulbs

                                                                       =588/600.

Coming to next question;

4(a) given,

Two dices are thrown simultaneously

So,

Total possible outcome = 6*6 = 36

(i)  The first dice shows 5

Favorable outcomes are

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

Probability will be = favorable/ total

                               = 6/36

                               = 1/6.

 

(ii)   Both the dices shows same number.

      Favorable outcomes are 

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

Probability will be = favorable/ total

                               = 6/36

                               = 1/6.

(iii)    The sum of number on the dice is 8

Favorable outcomes are 

(1,7),(2,6),(3,5),(4,4),(5,3),(6,2)=6

Probability will be = favorable/ total

                               = 6/36

                               = 1/6.

(iv)  the product of the numbers on the dice is 8

Favorable outcomes are 

(1,8),(2,4),(4,2),(8,1) = 4

 

Probability will be = favorable/ total

                               = 4/36

                               = 1/9.

(v)    The total number on the dice is grater than 9

  Favorable outcomes are 

(6,6),(6,5),(5,6),(5,5)= 4

Probability will be = favorable/ total

                               = 4/36

                               = 1/9.

Now come to next one!!!

5(a);- two coins are tossed once.

  Total possible outcome are (H,T),(T,H),(H,H),(T,T) = 4

(i)     Both tails

Favorable outcome is only one ;- (T,T)

Probability will be = favorable/ total

                               = 1/4.

(ii)   Atleast one head  

Favorable outcomes are

(H,T),(T,H),(H,H)= 3

Probability will be = favorable/ total

                               = 3/4.

(iii) Both head or both tail

Similarly to first one

Probability of getting both head =1/4

Probability of getting both tail  =1/4

Probability of getting both head or both tail = ¼ + ¼ = 2/4= ¼

5(b) Total no. of cards = 52

Total no. of face cards =12

Probability of drawing a face card =12/52= 3/13.


Anonymous: thnx for the help
Shubhendu8898: pleasure to help!:-)
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