There are the combinations that result in a total less than 7 and at least one die showing a 3:
[3, 3] [3,2] [2,1] [1,3] [2,3]
The probability of each of these is 1/6 * 1/6 = 1/36
There is a little ambiguity here about whether or not we should count [3,3] as the problem says "and one die shows a 3." Does this mean that only one die shows a 3 or at least one die shows a 3? Assuming the latter, the total probability is the sum of the individual probabilities:
1/36 + 1/36 + 1/36 + 1/36 + 1/36 = 5/36
Therefore, the required probability is: 5/36
Answer:
(D) P(AUB) = 0.65
Step-by-step explanation:
Since A and B are independent:
P(A ^ B)
= P(A) × P(B)
= 0.3 × 0.5 = 0.15
P(AUB) = P(A) + P(B) - P(A^B)
= 0.3 + 0.5 - 0.15
= 0.65
Answer:
Step-by-step explanation:
Step-by-step explanation:
answer is in photo above
