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3 years ago

A small deck of 5 cards consists of 3 red Jards and 2 green cards.

Mathematics

1 answer:

Simora [160]3 years ago

5 0

Answer:

Approximately $0.19$ .

Step-by-step explanation:

The cards were drawn with replacement. Therefore, the probability of drawing a red card will be $(3 / 5)$ at every one of the seven draws. The probability of not drawing a red card would be $(1 - (3/5)) = (2/5)$ .

One way to get three red cards out of the seven draws would be $\verb!red!\; \verb!red!\; \verb!red!\; \verb!green!\; \verb!green!\; \verb!green!\; \verb!green!$ . That corresponds to a probability of $(3/5)^3\, (2/5)^{7-3}$ .

However, there are many other ways to get three red cards during the seven draws. For example:

$\verb!red!\; \verb!red!\; \verb!green!\; \verb!red!\; \verb!green!\; \verb!green!\; \verb!green!$ ,
$\verb!red!\; \verb!red!\; \verb!green!\; \verb!green!\; \verb!red!\; \verb!green!\; \verb!green!$ .
$\cdots$ .
$\verb!green!\; \verb!green!\; \verb!green!\; \verb!green!\; \verb!red!\; \verb!red!\; \verb!red!$ .

There are $\displaystyle {\genfrac{(}{)}{0}{0}{7}{3}} = 35$ such ways in total to choose three draws out of seven draws.

These ways are mutually-exclusive. Each of these ways has a probability of $(3/5)^3\, (2/5)^{7-3}$ . The probability for getting a red card on exactly three out of the seven draws would be:

$\displaystyle {\genfrac{(}{)}{0}{0}{7}{3}} \, (3/5)^3\, (1 - (3/5))^{7-3} \approx 0.19$ .

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What is the formula for the expected number of successes in a binomial experiment with n trials and probability of success p? C

charle [14.2K]

Answer:

$(D)E[ X ] =np.$

Step-by-step explanation:

Given a binomial experiment with n trials and probability of success p,

$f(x)=\left(\begin{array}{c}n\\k\end{array}\right)p^x(1-p)^{n-x}, 0\leq x\leq n$

$E(X)=\sum_{x=0}^{n}xf(x)= \sum_{x=0}^{n}x\left(\begin{array}{c}n\\k\end{array}\right)p^x(1-p)^{n-x}$

Since each term of the summation is multiplied by x, the value of the term corresponding to x = 0 will be 0. Therefore the expected value becomes:

$E(X)=\sum_{x=1}^{n}x\left(\begin{array}{c}n\\x\end{array}\right)p^x(1-p)^{n-x}$

Now,

$x\left(\begin{array}{c}n\\x\end{array}\right)= \frac{xn!}{x!(n-x)!}=\frac{n!}{(x-)!(n-x)!}=\frac{n(n-1)!}{(x-1)!((n-1)-(x-1))!}=n\left(\begin{array}{c}n-1\\x-1\end{array}\right)$

Substituting,

$E(X)=\sum_{x=1}^{n}n\left(\begin{array}{c}n-1\\x-1\end{array}\right)p^x(1-p)^{n-x}$

Factoring out the n and one p from the above expression:

$E(X)=np\sum_{x=1}^{n}n\left(\begin{array}{c}n-1\\x-1\end{array}\right)p^{x-1}(1-p)^{(n-1)-(x-1)}$

Representing k=x-1 in the above gives us:

$E(X)=np\sum_{k=0}^{n}n\left(\begin{array}{c}n-1\\k\end{array}\right)p^{k}(1-p)^{(n-1)-k}$

This can then be written by the Binomial Formula as:

$E[ X ] = (np) (p +(1 - p))^{n -1 }= np.$

5 0

4 years ago

What association does the scatter plot show

Verizon [17]

C. No Association
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3 0

3 years ago

Read 2 more answers

In a city, the distance between the library and the police station is 3 miles less than twice the distance between the police

Lyrx [107]

Answer:

Option 3 - 4 miles

Step-by-step explanation:

Given : In a city, the distance between the library and the police station is 3 miles less than twice the distance between the police station and the fire station. The distance between the library and the police station is 5 miles.

To find : How far apart are the police station and the fire station?

Solution :

Let the distance between police station and fire station be 'x'.

The distance between the library and the police station is 3 miles less than twice the distance between the police station and the fire station.

i.e. $2x-3$