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Natasha_Volkova [10]
4 years ago
10

A skilled chess player believes that when they play a novice opponent, there is a 90% probability they will be able to beat them

. let X = the number of times this player would win against 15 novice opponents.
Then the random variable x follows what type of distribution? (Select all that apply.)

a) Geometric
b) Binomial
c) Poisson
Mathematics
2 answers:
zavuch27 [327]4 years ago
7 0

Answer:

b) Binomial

c) Poisson

Step-by-step explanation:

The geometric distribution is the number of trials required to have r successes. The measures the number of sucesses(wins), not the number of trials required to win r games. So the geometric distribution does not apply.

For each match, there are only two possible outcomes, either the skilled player wins, or he does not. The probability of the skilled player winning a game is independent of other games. So the binomial distribution applies.

We can also find the expected number of wins of the skilled player, which is 15*0.9 = 13.5. The Poisson distribution is a discrete distribution in which the only parameter is the expected number of sucesses. So the Poisson distribution applies.

So the correct answer is:

b) Binomial

c) Poisson

luda_lava [24]4 years ago
4 0

Answer:

Binomial

Step-by-step explanation:

No. of trials finite,

Success/failure

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First of all, note that all integers are either 0,1, or 2 modulo 3 (if you're not familiar with this terminology, it means that every integer is either a multiple of 3, or it is 1 or 2 away from a multiple of 3).

So, we can think of our numbers as

\begin{array}{c|c}x&x\mod 3\\0&0\\1&1\\2&2\\3&0\\4&1\\5&2\\6&0\\7&1\\8&2\\9&0\end{array}

In order to make sure that the sum of any three adjacent numbers is divisible by 3, we have to make sure that any group of 3 three adjacent numbers contains a 0, a 1 and a 2. This is possible only if we arrange our 9 numbers in 3 groups of 3 numbers containing 0,1 and 2 exactly once, repeating always the same pattern.

For example, we could arrange our numbers following the pattern

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216\cdot 6 = 1296

possible arrangements.

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