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masha68 [24]
3 years ago
12

You are playing a board game with your little sister. Moves are determined by rolling 2 six-sided dice. The red die tells you th

e direction of your move, and has 4 faces
that say "forward and 2 faces that say backward." The green die tells you how far to move, and has 1. 1. 2.2.3, and 4 on the faces. On your turn, what is the
probability that you move backward 2 spaces
Mathematics
1 answer:
enot [183]3 years ago
7 0

Answer: The probability is 1/9.

Step-by-step explanation:

First, let's define the possible outcomes of each dice:

Red: Forward (4 times), Backward (2 times)

Green : {1, 1, 2, 2, 3, 4}

We want to find the probability of moving backward 2 spaces.

Then we need to find the probability of rolling a "backward" in the red dice, and a 2 in the green dice.

First, the probability of rolling a backward in the red dice is equal to the quotient between the number of outcomes that are "backward", and the total number of outcomes in the dice (there are 2 backwards and 6 outcomes in total), this is:

p1 = 2/6 = 1/3.

And the probability of rolling a 2 in the green dice is equal to the quotient between the number of outcomes with a 2, and the total number of outcomes. (The 2 appears two times, and there are 6 possible outcomes):

p2 = 2/6 = 1/3.

Now, the probability of both events happening at the same time is equal to the product of the individual probabilities, then the probability of moving backwards 2 spaces is:

P = p1*p2 = (1/3)*(1/3) = 1/9.

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Suppose that 37% of college students own cats. If you were to ask random college students if they own a cat what would the proba
Likurg_2 [28]

Using the binomial distribution, the probabilities are given as follows:

a) 0.37 = 37%.

b) 0.5065 = 50.65%.

c) 0.3260 = 32.60%.

<h3>What is the binomial distribution formula?</h3>

The formula is:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

For this problem, the fixed parameter is:

p = 0.37.

Item a:

The probability is P(X = 1) when n = 1, hence:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 1) = C_{1,1}.(0.37)^{1}.(0.63)^{0} = 0.37

Item b:

The probability is P(X = 3) when n = 3, hence:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 3) = C_{3,3}.(0.37)^{3}.(0.63)^{0} = 0.5065

Item c:

The probability is P(X = 2) when n = 4, hence:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 2) = C_{4,2}.(0.37)^{2}.(0.63)^{2} = 0.3260

More can be learned about the binomial distribution at brainly.com/question/24863377

#SPJ1

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