Question

4.
An unbiased cubical dice has six faces numbered 4, 6,10,12,15, and 24. The die is thrown
twice and the Highest common factor of both results is recorded.
(a) Draw a possibility diagram (Sample space diagram) to show the possible outcomes
(b) Calculate the probability that:
(0) The HCF is 5
(ii) The HCF is greater than 3
(iii) The HCF is not 12
(iv) The HCF is not 24
(v) The HCF is even number
(vi) The HCF is equal to only one of the two numbers thrown.

Answer 1

(a) The sample space consists of 36 outcomes.

They are

{(4,4),(4,6),(4,10),(4,12),(4,15),(4,24),

(6,4),(6,6),(6,10),(6,12),(6,15),(6,24),

(10,4),(10,6),(10,10),(10,12),(10,15),(10,24),

(12,4),(12,6),(12,10),(12,12),(12,15),(12,24),

(15,4),(15,6),(15,10),(15,12),(15,15),(15,24),(24,4),(24,6),(24,10),(24,12),(24,15),(24,24)}

Each of these elements have the probability 1/36

(b)

(0) Elements having HCF = 5 are (10,15) and (15,10).

Probability of HCF is 5 = P(10,15) + P(15,10)

= 1/36 + 1/36

= 2/36

= 1/18

(ii) Elements having HCF greater than 3 are (4,4),(4,12),(4,24),(6,6),(6,12),(6,24),(10,10),(10,15),(12,4),(12,6),(12,12),(12,24),(15,10),(15,15),(24,4),(24,6),(24,12) and (24,24).

There are 18 elements.

Probability of HCF greater than 3 = 18/36

= 1/2

(iii) Elements having HCF = 12 are (12,12),(12,24) and (24,12).

Probabilty of HCF is 12 = 3/36

= 1/12

Probability of HCF not 12 = 1 - 1/12

= 11/12

(iv) Elements having HCF = 24 are (24,24)

Probability of HCF is 24 = 1/36

Probability of HCF not 24 = 1 - 1/36

= 35/36

(v) Elements having even HCF are (4,4),(4,6),(4,10),(4,12),(4,24),(6,4),(6,6),(6,10),(6,12),),(6,24),(10,4),(10,6),(10,10),(10,12),(10,24),(12,4),(12,6),(12,10),(12,12),(12,24),(24,4),(24,6),(24,10),(24,12) and (24,24).

There are 25 elements.

Probability of even HCF = 25/36

(vi) Elements with HCF equal to one of the numbers are (4,4),(4,12),(4,24),(6,6),(6,12),(6,24),(10,10),(12,4),(12,6),(12,12),(12,24),(15,15),(24,4),(24,6),(24,12) and (24,24).

There are 16 elements.

Probabilty of HCF as one of the thrown numbers = 16/36

= 4/9

#SPJ3

Answer 2

Answer:

a) 36

b)

1) $\frac{1}{18}$

2) $\frac{1}{2}$

3) $\frac{11}{12}$

4) $\frac{35}{36}$

5) $\frac{25}{36}$

6) $\frac{4}{9}$

Step-by-step explanation:

1. The sample space consists of 36 outcomes.
   {(4,4),(4,6),(4,10),(4,12),(4,15),(4,24),(6,4),(6,6),(6,10),(6,12),(6,15),(6,24),(10,4),(10,6),(10,10),(10,12),(10,15),(10,24),(12,4),(12,6),(12,10),(12,12),(12,15),(12,24),(15,4),(15,6),(15,10),(15,12),(15,15),(15,24),(24,4),(24,6),(24,10),(24,12),(24,15),(24,24)}
2. Probablity of that numbers have HCF of 5 are (10,15) and (15,10)
   So probablity of getting them are $\frac{1}{36}$ each so
   $\frac{1}{36}+\frac{1}{36} =\frac{1}{18}$
3. Probablity of that numbers have HCF greater then 3 are (4,4),(4,12),(4,24),(6,6),(6,12),(6,24),(10,10),(10,15),(12,4),(12,6),(12,12),(12,24),(15,10),(15,15),(24,4),(24,6),(24,12) and (24,24).
   As there are 18 elements and each has a probablity of $\frac{1}{36}$ we get the probablity as $18 * \frac{1}{36} = \frac{1}{18}$
4. Probablity of that numbers do not have HCF of 24 are every thing except (24,24) so the probablity becomes $\frac{35}{36}$
5. The probablity of number that do not have HCF of 12 are everything except (12,12)(24,12)(12,24) hence the probablity becomes $\frac{33}{36}$ that is $\frac{11}{12}$
6. The HCF is even number
   (4,4),(4,6),(4,10),(4,12),(4,24),(6,4),(6,6),(6,10),(6,12),),(6,24),(10,4),(10,6),(10,10),(10,12),(10,24),(12,4),(12,6),(12,10),(12,12),(12,24),(24,4),(24,6),(24,10),(24,12) and (24,24).
   Hence that are 25 so prabablity becomes $\frac{25}{36}$
7. Elements with HCF equal to one of the numbers are (4,4),(4,12),(4,24),(6,6),(6,12),(6,24),(10,10),(12,4),(12,6),(12,12),(12,24),(15,15),(24,4),(24,6),(24,12) and (24,24) there are 16 elements so probablity is $\frac{16}{36}=\frac{4}{9}$
   #SPJ2