Question

die is thrown twice. Find the probability of getting : (i) both scores even (ii) both scores are

prime numbers (iii) sum of scores less than 5​

Answer 1

Answer:

$i) 1/4 \:,\: ii) 2/9\:,iii) 1/6$

Step-by-step explanation:

$\huge \text{Given}$

A die is thrown twice.

$\huge \text{To Find Probability of getting :-}$

(i) both scores even .

(ii) both scores are prime numbers .

(iii) sum of scores less than 5​ .

$\huge \text{Solution }$

Total Number of outcomes = 6 × 6 = 36 .

$\bigstar$ Sample Space when a die is thrown twice

(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 =               Number of favourable events              

                                                       Total number of Events

Probability That :-

i) Both scores are even

Favourable events are - (2,2) , (2,4) , (2,6) ,  (4,2) , (4,4) , (4,6) , (6,2) , (6,4) , (6,6)

Number of Favourable  events = 9

Total number of events = 36

So , The probability of getting both scores even = $\dfrac{9}\dfrac{}36$    =  $\dfrac{1}4$

(ii) both scores are prime numbers

Favourable events = (2,2) , (2,3) , (2,5) , (3,2) , (3,3) ,(3,5) , (5,3) , (5,5)

Number of Favourable events = 8

Probability = $\dfrac{8}\dfrac{36}$  = $\dfrac{2}9$

(iii) sum of scores less than 5​

Favourable events =  (1,1), (1,2), (1,3), (2,1), (2,2) , (3,1)

Number of Favourable events = 6

Probability = $\dfrac{6}\dfrac{36}$   = $\dfrac{1}6$