Even number can be either 2, 4 or 6. There is 1/6 chance for each one.
If previous was a 2, a 5 would = 7 (1/6 * 1/6) = 1/36
If previous was a 4, a 3 would = 7 (1/6 * 1/6) = 1/36
If previous was a 6, a 1 would = 7 (1/6 * 1/6) = 1/36
Either-or probability is found by multiplying the individual probabilities, 3/36.
What is the probability oi rolling a total of 7 in two rolls of a standard die if you get an even number of the first roll?
Could someone please help me? Thanks
3 answers
you roll a number cube, what are the odds in rolling a 1 or 4?
The question needs to be understood in a slightly different way. It is 'given' that the first roll was an even number. Hence the probability of a 2 is 1/3, for 4 is 1/3 and for 6 it is 1/3
If previous was a 2, a 5 would = 7-> 1/3 * 1/6 = 1/18
If previous was a 4, a 3 would = 7 -> 1/3 * 1/6 = 1/18
If previous was a 6, a 1 would = 7 -> 1/3 * 1/6 = 1/18
Any one of the above possibilities will result in desired outcome. It is an 'OR' situation. Hence add the possibilities to get 3/18 = 1/6
If previous was a 2, a 5 would = 7-> 1/3 * 1/6 = 1/18
If previous was a 4, a 3 would = 7 -> 1/3 * 1/6 = 1/18
If previous was a 6, a 1 would = 7 -> 1/3 * 1/6 = 1/18
Any one of the above possibilities will result in desired outcome. It is an 'OR' situation. Hence add the possibilities to get 3/18 = 1/6
1 answer