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1 year ago

If you select one card at random from a standard deck of 52 cards, what is the probability that the card is a king AND a spade?

Mathematics

1 answer:

ExtremeBDS [4]1 year ago

7 0

Given:

There are 52 cards.

So, n(s)=52

To find the probability that the card is a king and a spade:

Let A be an event of choosing a card that is king and a spade.

So, n(A)=1

Hence, the probability that the card is a king and a spade is,

$\frac{1}{52}$

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According to an article, 47% of adults have experienced a breakup at least once during the last 10 years. Of 9 randomly select

In-s [12.5K]

Answer:

a) $P(X=5)=(9C5)(0.47)^5 (1-0.47)^{9-5}=0.228$

b) $P(X=0)=(9C0)(0.47)^0 (1-0.47)^{9-0}=0.0033$

And replacing we got:

$P(X \geq 1)= 1-P(X$

c) $P(X=4)=(9C4)(0.47)^4 (1-0.47)^{9-4}=0.257$

$P(X=5)=(9C5)(0.47)^5 (1-0.47)^{9-5}=0.228$

$P(X=6)=(9C6)(0.47)^6 (1-0.47)^{9-6}=0.135$

And adding we got:

$P(4 \leq X \leq 6)= 0.257+0.228+0.135=0.620$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=9, p=0.47)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Assuming the following questions:

a. exactly five

For this case we can use the probability mass function and we got:

$P(X=5)=(9C5)(0.47)^5 (1-0.47)^{9-5}=0.228$

b. at least one

For this case we want this probability:

$P(X \geq 1)$

And we can use the complement rule and we got:

$P(X \geq 1)= 1-P(X$

$P(X=0)=(9C0)(0.47)^0 (1-0.47)^{9-0}=0.0033$

And replacing we got:

$P(X \geq 1)= 1-P(X$

c. between four and six, inclusive.

For this case we want this probability:

$P(4 \leq X \leq 6)$

$P(X=4)=(9C4)(0.47)^4 (1-0.47)^{9-4}=0.257$

$P(X=5)=(9C5)(0.47)^5 (1-0.47)^{9-5}=0.228$

$P(X=6)=(9C6)(0.47)^6 (1-0.47)^{9-6}=0.135$

And adding we got:

$P(4 \leq X \leq 6)= 0.257+0.228+0.135=0.620$

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3 years ago

An excellent swimmer is confident he can judge his speed in the water exactly. He bets he can swim two lengths of a 50-meter Oly

Salsk061 [2.6K]

The time needed at 2 meters per second for both rounds would be 100/2 s or 50 seconds. For the first lane he already used 50/1.5 -> 33.33 seconds, so he would need to cover the other 50 meters in exactly 16.67. 50m divided by the 16.67 seconds he has left means he has to swim at a speed of 3 meters per second.

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3 years ago

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Answer:

see below

Step-by-step explanation:

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ie. question 6

log ₁₀(3x+1) = 2 -----> $10^{2} = 3x+1$

that will get you through questions 1 to 3, 5 to 6, and 8

in question 4, all you have to do is know that 2^2 = 4 and 2^3 = 8, by setting the bases equal, you can manipulate the exponents to get 2x+8 = 3x-3

for questions 7 and 9,

remember that:

logₐc + logₐd = logₐ(cd)

logₐc - logₐd = logₐ( $\frac{c}{d}$ )

remember change of base is $log_{a} b = \frac{log_xb}{log_xa}$ , this will be useful if you need your calculator since calculators only have base 10 and maybe if your calculator is good enough natural base e