- #1
jdstokes
- 523
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A six-sided die is loaded in such a way that an odd number is twice as likely to occur as an even number. We throw the die until a 3 is observed. What is the probability that the number of throws required is (strictly) more than 5 and (strictly) less than 10?
Answer: 0.1805.
My working.
P(odd) = 2P(even)
P(odd) + P(even) = 1
Therefore
P(odd) = 2/9.
X = "number of throws before success"
X is geometric with p = 2/9.
Therefore
[itex]P( 6\leq X \leq 9 ) = \sum_{i=6}^9 (2/9)(1-2/9)^i = 0.1403642 [/itex].
Any help would be appreciated.
Thanks.
James
Answer: 0.1805.
My working.
P(odd) = 2P(even)
P(odd) + P(even) = 1
Therefore
P(odd) = 2/9.
X = "number of throws before success"
X is geometric with p = 2/9.
Therefore
[itex]P( 6\leq X \leq 9 ) = \sum_{i=6}^9 (2/9)(1-2/9)^i = 0.1403642 [/itex].
Any help would be appreciated.
Thanks.
James