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Marina86 [1]
2 years ago
5

Your friend has two standard decks of 52 playing cards and asks you to randomly draw one card from each deck. what is the probab

ility that you will draw two spades? express your first answer as a fraction in simplest form, and round your percent answer to the nearest hundredth.
Mathematics
2 answers:
Mila [183]2 years ago
5 0
There are 13 spades in one deck, so the probability to draw a spade from the first deck is 
P_1=\frac{1}{4}

Similarly, the probability to draw a spade from the second deck is
P_2=\frac{1}{4}

The probability that both events occur (so, the probability we have to draw two spades from the two decks) is the product of the two probabilities, so:
P=P_1 \cdot P_2 =  \frac{1}{4} \cdot  \frac{1}{4}= \frac{1}{16}
Nimfa-mama [501]2 years ago
3 0

Answer:

The probability that you would draw two spades are 1/16

Step-by-step explanation:

Your friend has two standard decks of 52 playing cards and asks you to randomly draw one card from each deck.

The probability that you would draw both are spades.

We know that there are 13 spades in a deck of 52 cards. so the probability of first draw would be spade is

P1  = \frac{13}{52}= \frac{1}{4}

From another deck you draw second card so the probability of second draw would be spade is

P2 = \frac{13}{52}= \frac{1}{4}

Therefore, the probability of drawing two spades is

P = \frac{1}{4} ×  \frac{1}{4} = \frac{1}{16}

The probability that you would draw two spades are 1/16

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The equation of the line would be y = (1/2)x.

To find the equation of a line that is reflected through the line y = x, we just switch the x and y. Then, solve for y.

y = 2x
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2 years ago
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Oduvanchick [21]

Answer:

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Hope it's right:)

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2 years ago
Triangle XYZ has vertices X(–1, –1), Y(–2, 1), and Z(1, 2). What is the approximate measure of angle Z?
Molodets [167]
The answer to this question is A. 37.2°
8 0
2 years ago
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The factory quality control department discovers that the conditional probability of making a manufacturing mistake in its preci
Anni [7]

Answer:

The probability that a defective ball bearing was manufactured on a Friday = 0.375

Step-by-step explanation:

Let the event of making a mistake = M

The event of making a precision ball bearing production on Monday = Mo

The event of making a precision ball bearing production on Tuesday = T

The event of making a precision ball bearing production on Wednesday = W

The event of making a precision ball bearing production on Thursday = Th

The event of making a precision ball bearing production on Friday = F

the conditional probability of making a manufacturing mistake in its precision ball bearing production is 4% on Tuesday, P(M|T) = 4% = 0.04

4% on Wednesday, P(M|W) = 0.04

4% on Thursday, P(M|Th) = 0.04

8% on Monday, P(M|Mo) = 0.08

and 12% on Friday = P(M|F) = 0.12

The Company manufactures an equal amount of ball bearings (20 %) on each weekday, Hence, the probability that a random precision ball bearing was made on a particular day of the week, is mostly the same for all the five working days.

P(Mo) = 0.20

P(T) = 0.20

P(W) = 0.20

P(Th) = 0.20

P(F) 0.20

The probability that a defective ball bearing was manufactured on a Friday = P(F|M)

P(F|M) = P(F n M) ÷ P(M)

P(F n M) = P(M n F)

P(M) = P(Mo n M) + P(T n M) + P(W n M) + P(Th n M) + P(F n M)

We can obtain each of these probabilities by using the expression for conditional probability.

P(Mo n M) = P(M|Mo) × P(Mo) = 0.08 × 0.20 = 0.016

P(T n M) = P(M|T) × P(T) = 0.04 × 0.20 = 0.008

P(W n M) = P(M|W) × P(W) = 0.04 × 0.20 = 0.008

P(Th n M) = P(M|Th) × P(Th) = 0.04 × 0.20 = 0.008

P(F n M) = P(M|F) × P(F) = 0.12 × 0.20 = 0.024

P(M) = P(Mo n M) + P(T n M) + P(W n M) + P(Th n M) + P(F n M)

P(M) = 0.016 + 0.008 + 0.008 + 0 008 + 0.024 = 0.064

P(F|M) = P(F n M) ÷ P(M)

P(F n M) = P(M n F) = 0.024

P(M) = 0.064

P(F|M) = P(F n M) ÷ P(M) = (0.024/0.064) = 0.375

Hope this Helps!

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