Question

Find the conditional probability, in a single roll of two fair 6-sided dice, that the sum is less than 6, given that the sum is even The probability is 2 (Type an integer or a simplified fraction) 1

Answer 1

Hello

YOUR REQUIRED ANSWER IS 2/9

	1	2	3	4	5	6
1	2		4		6
2		4		6		8
3	4		6		8
4		6		8		10
5	6		8		10
6		8		10		12

$The \ required \ probability =\frac{4}{18}=\frac{2}{9}$

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Answer 2

SOLUTION :

Events of even sum of two dice after rolling : 

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

(6, 6)  : 

Total 18 events 

Probability of each event = 1/6 * 1/6 = 1/36 

P(even sum) = 18 * 1/36 = 1/2.

Events of sum  less than 6 :   (1, 1), (1, 2), (2, 1), (1, 3), (3, 1), (2, 2), (1, 4), (4, 1), (2, 3), (3, 2)  

Total 10 events 

Probability of each event = 1/6 * 1/6 = 1/36 

P(sum < 6) = 10 * 1/36 = 5/18 .

Events : (1, 1), (1, 3), (3, 1) and (2, 2) are common to both cases,

=> P(sum even AND sum < 6) : 4 * 1/36 = 1/9 

So,

P( sum < 6 | sum even) 

= P(sum even AND sum < 6) / P(sum even)

= (1/9) / (1/2)

= 2/9 (ANSWER).

ALTERNATELY ;

P( sum < 6 | sum even)  

= Common events / Space of even events

= 4 / 18 

= 2/ 9 (ANSWER).