Probability

The newest invention of the 6.041x staff is a three-sided die. On any roll of this die, the result is 1 with probability 1/2, 2 with probability 1/4, and 3 with probability 1/4.

Consider a sequence of six independent rolls of this die.

1. Find the probability that exactly two of the rolls results in a 3.

2. Given that exactly two of the six rolls resulted in a 1, find the probability that the first roll resulted in a 1.

3. We are told that exactly three of the rolls resulted in a 1 and exactly three rolls resulted in a 2. Given this information, find the probability that the six rolls resulted in the sequence (1,2,1,2,1,2).

4. The conditional probability that exactly k rolls resulted in a 3, given that at least one roll resulted in a 3, is of the form:
11−(c1/c2)c3(c3k)(1c2)k(c1c2)c3−k,for k=1,2,…,6.

Find the values of the constants c1, c2, and c3:

1. 👍 0
2. 👎 0
3. 👁 516
1. part-2
ans:= 1/3

1. 👍 0
2. 👎 0
2. 1. (6 2)*(1/4)^2*(3/4)^4

3. 0.05

4. c1=3, c2= 4, c3=6

1. 👍 0
2. 👎 0
posted by RVE

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