First, here's a restatement of yesterday's probability question :
Compare these 2 probabilities and explain method:
(a) Prob of rolling exactly 3 sixes in 5 rolls of a fair die.
(b) Prob of rolling exactly 3 sevens in 5 rolls of a pair of fair dice
Discussion :
Both are examples of binomial probability because they involve repeated independent trials each of which has 2 outcomes. The following explanation is intentionally detailed and 'repetitious'.
The prob of a 6 on each roll is 1/6. Each roll produces only 2 outcomes, either a 6 (prob=1/6) or not a 6 (prob = 5/6).
The prob of a 7 on each roll of a pair of dice is 6/36 or 1/6. Each roll of the pair has only 2 outcomes, either a 7 (prob=1/6) or not a 7 (prob=5/6).
Therefore, the probabilities of getting 3 successes in 5 trials is the same. Since the question asks for a comparison, we're done.
The actual prob is C(5,3)(1/6)^3•(5/6)^2 where C(5,3) is the 'MathNotation' for the number of ways of arranging 5 objects, one group of 3 identical objects and a separate group of 2 identical objects. This is not the usual way of defining combinations but I like this interpretation.
I guess the QuadAnagram was a bit challenging. Here's a hint for the ending:
...he's a bored L---R.
Email me at dmarain at gmail dot com with your answer.
Sent from my Verizon Wireless 4GLTE Phone
Compare these 2 probabilities and explain method:
(a) Prob of rolling exactly 3 sixes in 5 rolls of a fair die.
(b) Prob of rolling exactly 3 sevens in 5 rolls of a pair of fair dice
Discussion :
Both are examples of binomial probability because they involve repeated independent trials each of which has 2 outcomes. The following explanation is intentionally detailed and 'repetitious'.
The prob of a 6 on each roll is 1/6. Each roll produces only 2 outcomes, either a 6 (prob=1/6) or not a 6 (prob = 5/6).
The prob of a 7 on each roll of a pair of dice is 6/36 or 1/6. Each roll of the pair has only 2 outcomes, either a 7 (prob=1/6) or not a 7 (prob=5/6).
Therefore, the probabilities of getting 3 successes in 5 trials is the same. Since the question asks for a comparison, we're done.
The actual prob is C(5,3)(1/6)^3•(5/6)^2 where C(5,3) is the 'MathNotation' for the number of ways of arranging 5 objects, one group of 3 identical objects and a separate group of 2 identical objects. This is not the usual way of defining combinations but I like this interpretation.
I guess the QuadAnagram was a bit challenging. Here's a hint for the ending:
...he's a bored L---R.
Email me at dmarain at gmail dot com with your answer.
Sent from my Verizon Wireless 4GLTE Phone
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