The sample space for the experiment N(S) = 12 and the probability of showing 2 on at least one die is 41/81.
Probability is the chance of occurrence of a certain event out of the total no. of events that can occur in a given context.
Given that four three-sided dice(4, 5, 6 are replaced by 1, 2, 3) are rolled.
As each die has two 1's, two 2's, and two 3's the no. of distinct sample space N(S) each die has is 3, and there are a total of four dice so the total no.of sample space is (3×4) = 12.
∴ The probability that one of the die is showing 2 is (2/6) = 1/3.
So, the probability that at least one of the die is showing two is sum of one die showing two to sum of 4 dice showing 2 which is,
= 2/6 + (2/6)(2/6) + (2/6)(2/6)(2/6) + (2/6)(2/6)(2/6)(2/6).
= 1/3 + 1/9 + 1/27 + 1/81.
= (27 + 9 + 3 + 1)/81.
= 40/81.
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Answer: so the rest is 7, 14-7 = 7.
Step-by-step explanation:
A. -9w + 4
B. -9w – 4
C. w + 4
D. w – 4
Answer:
Step-by-step explanation:
D. W - 4
Answer:
The answer is D. W-4
Step-by-step explanation:
Answer:8300
Step-by-step explanation: