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Sliva [168]
3 years ago
6

A card is drawn from a normal 52 card deck. The card number and sum are noted, the card is placed back in the deck, then another

card is drawn. Find the probability that a) the first card was a queen and the second card as a 10. b) the first card was the Queen of Hearts and the second card was an 8. c) both cards were Jacks. d) the first card was the 5 of Spades and the second card was the Ace of Hearts.
Mathematics
1 answer:
Solnce55 [7]3 years ago
3 0

Step-by-step explanation:

a) There are 4 Queens in a deck, so the probability the first card is a Queen is 4/52 = 1/13.

There are 4 10s in a deck, so the probability the second card is a 10 is 4/52 = 1/13.

The probability of both events is 1/13 × 1/13 = 1/169.

b) There is 1 Queen of Hearts in a deck, so the probability the first card is a Queen of Hearts is 1/52.

There are 4 8s in a deck, so the probability the second card is a 8 is 4/52 = 1/13.

The probability of both events is 1/52 × 1/13 = 1/676.

c) There are 4 Jacks in a deck, so the probability the first card is a Jack is 4/52 = 1/13.

There are 4 Jacks in a deck, so the probability the second card is a Jack is 4/52 = 1/13.

The probability of both events is 1/13 × 1/13 = 1/169.

d) There is 1 5 of Spades in a deck, so the probability the first card is a 5 of Spades is 1/52.

There is 1 Ace of Hearts in a deck, so the probability the first card is a Ace of Hearts is 1/52.

The probability of both events is 1/52 × 1/52 = 1/2704.

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3 years ago
A package delivery service claims that no more than 5 percent of all packages arrive at the address late. Assuming that the cond
masya89 [10]

Answer: 0.0115

Step-by-step explanation:

The binomial probability formula :-

P(X)=^nC_x\ p^x(1-p)^{n-x}, here n is the number total of trials , p is the probability of getting success in each trial and P(x) is the probability of getting success in x trial.

Given : A package delivery service claims that no more than 5 percent of all packages arrive at the address late.

We assume that the conditions for the binomial​ hold with parameters :-

n=10 ; p=0.05

Let x be the random variable that represents the number of packages arrive at the address late.

Now, the probability that more than 2 packages will be delivered​ late :-

P(x>2)=1-P(x\leq2)\\\\=1-(P(0)+P(1)+P(2))\\\\=1-(^{10}C_0(0.05)^{0}(1-0.05)^{10}+^{10}C_{1}(0.05)^1(1-0.05)^9+^{10}C_{2}(0.05)^2(1-0.05)^8)\\\\=1-((0.95)^{10}+(10)(0.05)(0.95)^9+45(0.05)^2(0.95)^8}\\\\=1-0.988496442621=0.011503557379\approx0.0115

Hence, the probability that more than 2 packages will be delivered​ late = 0.9885

4 0
3 years ago
How do I solve this equation
notka56 [123]

Hello there!

\frac{x}{y}=h-k

Explanation:

↓↓↓↓↓↓↓↓

First you had to multiply by y from both sides of the equations.

\frac{xy}{y}=hy-ky; y\neq0

Simplify it should be the correct answer.

x=hy-ky; y\neq0

Answer⇒⇒⇒⇒⇒x=hy-ky; y≠0

Hope this helps!

Thank you for posting your question at here on Brainly.

Have a great day!

-Charlie

6 0
3 years ago
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