The rolls are independent. If you have already obtained a 1, you are asking for the probability of getting either a 1 or 2 to add to "less than 4."
1/6 + 1/6 = 2/6 = 1/3
With 6 possibilities for any roll of a die, there is no way of getting a denominator of 11.
http://www.members.cox.net/dagershaw/lol/Odds.html
A die is rolled twice. What is the probability that the sum of the rolls is less than 4 given that one of the rolls is a 1?
The answer is 3/11. Could someone tell me how to get this. Thanks?
2 answers
Possibilities with rolling 3 on ONE of the rolls:
First roll: 1,1 | 1,2 | 1,3 | 1,4 | 1,5 | 1,6
Second roll: 1,1 | 2,1 | 3,1 | 4,1 | 5,1 | 6,1
Eliminate 1,1 from second roll because it is the same as in the first, therefore there are 11 possibilities with 1 as either the first or second roll.
Find the rolls that are less than 4.
(1,1) (1,2) (2,1). 3 rolls are less than 4.
Therefore P(x < 4) = 3 / 11.
First roll: 1,1 | 1,2 | 1,3 | 1,4 | 1,5 | 1,6
Second roll: 1,1 | 2,1 | 3,1 | 4,1 | 5,1 | 6,1
Eliminate 1,1 from second roll because it is the same as in the first, therefore there are 11 possibilities with 1 as either the first or second roll.
Find the rolls that are less than 4.
(1,1) (1,2) (2,1). 3 rolls are less than 4.
Therefore P(x < 4) = 3 / 11.
2 answers