Find the probability of seeing at most 5 dots, but also an odd number of dots when 2 fair dice are rolled.
at most 5 dots --> 2 , 3, 4, 5 dots
11, 12, 21, 22, 23, 32 ---> 6 ways
showing odd number of dots: 3, 5, 7 , 9 , 11
{12, 21, 23, 32, 43, 34 , 54, 45 , 65 , 56}
so we want {at most 5 dots} AND {showing odd sum}
= {12, 21, 23, 32}
or 4 ways
prob(stated event) = 4/36 = 1/9
To find the probability of seeing at most 5 dots, but also an odd number of dots when rolling 2 fair dice, we need to consider two cases: the case where the total sum of the dots is odd and at most 5, and the case where the total sum is even but still at most 5.
Case 1: Total sum is odd and at most 5
In this case, the possible outcomes are (1, 1), (1, 3), (1, 5), (2, 1), (2, 3), (3, 1), (3, 5), (4, 1), (4, 3), (5, 1), (5, 3), (6, 1). There are 12 possible outcomes in total.
Case 2: Total sum is even and at most 5
In this case, the possible outcomes are (1, 2), (1, 4), (2, 1), (2, 3), (3, 2), (4, 1). There are 6 possible outcomes in total.
To get the probability, we need to divide the number of favorable outcomes by the total number of possible outcomes.
Total number of possible outcomes = 6 * 6 = 36
Number of favorable outcomes = 12 + 6 = 18
Probability = number of favorable outcomes / total number of possible outcomes
Probability = 18 / 36
Probability = 1/2 or 0.5
Therefore, the probability of seeing at most 5 dots, but also an odd number of dots when rolling 2 fair dice is 1/2 or 0.5.