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Nookie1986 [14]
3 years ago
10

A number cube is rolled and a coin is flipped.find the probabilities below

Mathematics
1 answer:
S_A_V [24]3 years ago
8 0

The two events are independent, because the result of the number cube has no influence on the coin flipped, and vice versa. So, we can calculate the two probabilities independently and then multiply them.

So, we have:

P(1 and heads) = P(cube returns 1) * P(coin returns heads)

The probability of rolling 1 is 1/6, because there are 6 numbers on the cube, and they all have the same probability. Similarly, the probability of getting heads is 1/2, because it's either heads or tails with equal probability.

P(\text{1 and heads}) = \dfrac{1}{6}\cdot\dfrac{1}{2}=\dfrac{1}{12}

The other two exercises follow the exact same logic, you only need to count how many outcomes satisfy your request for both the cube and the coin, and multiply the probabilities:

P(\text{even and tails})=\dfrac{1}{2}\cdot\dfrac{1}{2}=\dfrac{1}{4}

P(\text{5 or 6 and heads})=\dfrac{2}{6}\cdot\dfrac{1}{2}=\dfrac{1}{6}

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A chemical supply company currently has in stock 100 lb of a certain chemical, which it sells to customers in 5-lb batches. Let
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Complete Question:

A chemical supply company currently has in stock 100 lb of a certain chemical, which it sells to customers in 5-lb batches. Let X = the number of batches ordered by a randomly chosen customer, and suppose that X has the pmf shown in the table below. Compute E(X) and V(X).

Then compute the expected number of pounds left after the next customer’s order is shipped and the variance of the number of pounds left.

NOTE: The pmf is shown in the file attached

Answer:

Expected number of batches, E(X) = 2.0 batches

Variance of batches, V(X) = 0.8 batches²

Expected number of pounds left, E(Y) = 90 lb

Variance of pounds left, V(Y) = 20 lb²

Step-by-step explanation:

The company has 100 batches and it sells to customers in 5 lb batches. The number of pounds left after each customer is filled will be modeled by the mathematical formula: Y = 100 - 5X

The expected value of X, E(X) can be calculated using the formula:

E(X) = \sum x p(x)

E(X) = (1*0.3) + (2*0.5) + (3*0.1) + (4*0.1)

E(X) = 0.3 + 1 + 0.3 + 0.4

E(X) = 2.0

Expected number of batches, E(X) = 2.0 batches

The variance of X, V(X) can be calculated using the formula:

V(X) = E(X^2) - [E(X)]^2\\ E(X^2) = \sum x^2 p(x)\\  E(X^2) = (1^2 *0.3) + (2^2 * 0.5) + (3^2 * 0.1) + (4^2 * 0.1)\\  E(X^2) = 0.3 + 2 + 0.9 + 1.6\\ E(X^2) = 4.8\\V(X) = 4.8 - 2^2\\V(X) = 0.8

Variance of batches, V(X) = 0.8 batches²

Y = 100 - 5X

E(Y) = E(100 - 5X)

E(Y) = 100 - 5E(X)

E(Y) = 100 - 5(2)

E(Y) = 90 lb

Expected number of pounds left, E(Y) = 90 lb

Variance of the number of pounds left:

V(Y) = V(100 - 5X)

V(Y) = V(100) + V(-5X)

V(100) = 0

V(Y) = (-5)² V(X)

V(Y) = 25 * 0.8

V(Y) = 20

Variance of pounds left, V(Y) = 20 lb²

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Answer:

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so x= 25

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