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A 5-card hand is dealt from a perfectly shuffled deck. Define the events: A: the hand is a four of a kind (all four cards of one

TiliK225 [7]

In a hand of 5 cards, you want 4 of them to be of the same rank, and the fifth can be any of the remaining 48 cards. So if the rank of the 4-of-a-kind is fixed, there are $\binom44\binom{48}1=48$ possible hands. To account for any choice of rank, we choose 1 of the 13 possible ranks and multiply this count by $\binom{13}1=13$ . So there are 624 possible hands containing a 4-of-a-kind. Hence A occurs with probability

$\dfrac{\binom{13}1\binom44\binom{48}1}{\binom{52}5}=\dfrac{624}{2,598,960}\approx0.00024$

There are 4 aces in the deck. If exactly 1 occurs in the hand, the remaining 4 cards can be any of the remaining 48 non-ace cards, contributing $\binom41\binom{48}4=778,320$ possible hands. Exactly 2 aces are drawn in $\binom42\binom{48}3=103,776$ hands. And so on. This gives a total of

$\displaystyle\sum_{a=1}^4\binom4a\binom{48}{5-a}=886,656$

possible hands containing at least 1 ace, and hence B occurs with probability

$\dfrac{\sum\limits_{a=1}^4\binom4a\binom{48}{5-a}}{\binom{52}5}=\dfrac{18,472}{54,145}\approx0.3412$

The product of these probability is approximately 0.000082.

A and B are independent if the probability of both events occurring simultaneously is the same as the above probability, i.e. $P(A\cap B)=P(A)P(B)$ . This happens if

the hand has 4 aces and 1 non-ace, or
the hand has a non-ace 4-of-a-kind and 1 ace

The above "sub-events" are mutually exclusive and share no overlap. There are 48 possible non-aces to choose from, so the first sub-event consists of 48 possible hands. There are 12 non-ace 4-of-a-kinds and 4 choices of ace for the fifth card, so the second sub-event has a total of 12*4 = 48 possible hands. So $A\cap B$ consists of 96 possible hands, which occurs with probability

$\dfrac{96}{\binom{52}5}\approx0.0000369$

and so the events A and B are NOT independent.

4 0

3 years ago

Y=log x If y=10, then what is x?

netineya [11]

Answer:

First question, x= 10^10.

Second question is 10!. or 362880

Step-by-step explanation:

First Question:

Simple logx has a base of 10, i.e log10 x,

the question will be 10 = log10 x,

when taking the base "10" from the right side to the left, the number on the left side becomes the power of the base, in this case 10 from the right will be base and 10 from the left will power and log will vanish.

x=10^10.

Another example with different numbers

Y=logx if Y= 12, What is x?

The base is ten when not given,so:

12=logx

10^12=x

Second Question;

simple multiplication just multiply the numbers.

10! is pronounced as 10 factorial,

5! will be 5x4x3x2x1=120

8 0

3 years ago

Write the quadratic equation of the parabola that passes through (-2,5) and has a vertex of (1,-3)

guapka [62]

-8/3x - 1/3 is the equation

6 0

3 years ago

PLZ HELP I WILL GIVE THE BRAINLEST AND 22 PTS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! An expression is shown below: 6(m + 2 +

n200080 [17]

Answer:

part A: expression 1: 6(8m+2)

expression 2: 6m+12+42m

Part B: 6(m+2+7m)= 6(8m+2)

combine like terms is 6(m+2+7m), so it is 6(2+8m) or 6(8m+2)

6(8m+2)= 6(8m+2)

Part C: 6m+12+42m=6(m+2+7m)

m=0

6(0+2+7(0))=6(0)+12+42(0)

6(2)= 12

12=12

7 0

3 years ago

You are to take a multiple-choice exam consisting of 64 questions with 5 possible responses to each question. Suppose that you h

damaskus [11]

Answer:

a) Expected score on the exam is 12.8.

b) Variance 10.24, Standard deviation 3.2

Step-by-step explanation:

For each question, there are only two possible outcomes. Either you guesses the answer correctly, or you does not. The probability of guessing the answer of a question correctly is independent of other questions. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

The expected value of the binomial distribution is:

$E(X) = np$