Pleaseeeeeeeeee help meeee I don’t rlly get it

Imagine you have a classmate who has never had an introduction to statistics in a math course. Explain to that person how the pr

Tamiku [17]

Answer:

Probability of selecting red tile out of a box containing 3 red,2 white & 5 black tiles = 3/10 = 0.3

Step-by-step explanation:

Probability is the calculation of likelihood of an event.

It can be calculated by Formula : Favourable outcome / Total Outcomes

Example : If a box contain 3 red tiles, 2 white tiles, 5 black tiles

Prob (selecting a red tile) = Favourable outcome (red tiles) / Total Outcomes (total tiles) = 3 / 10 = 0.3

Similarly, prob (white tile) & prob (black tile) = 2/10 ie 0.2 & 5/10 ie 0.5 respectively

8 0

3 years ago

Addmaths probability. I need helpp for question a and b thank youu

motikmotik

Answer:

sorry this is soo blur

please follow me and mark has brilliant please

7 0

3 years ago

A bag contains 4 blue marbles and 2 yellow marbles. Two marbles are randomly chosen (the first marble is NOT replaced before dra

VMariaS [17]

Answer:

0.4 = 40% probability that both marbles are blue.

0.0667 = 6.67% probability that both marbles are yellow.

53.33% probability of one blue and then one yellow

If you are told that both selected marbles are the same color, 0.8571 = 85.71% probability that both are blue

Step-by-step explanation:

To solve this question, we need to understand conditional probability(for the final question) and the hypergeometric distribution(for the first three, because the balls are chosen without being replaced).

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

What is the probability that both marbles are blue?

4 + 2 = 6 total marbles, which means that $N = 6$

4 blue, which means that $k = 4$

Sample of 2, which means that $n = 2$

This is P(X = 2). So

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 2) = h(2,6,2,4) = \frac{C_{4,2}*C_{2,0}}{C_{6,2}} = 0.4$

0.4 = 40% probability that both marbles are blue

What is the probability that both marbles are yellow?

4 + 2 = 6 total marbles, which means that $N = 6$

2 yellow, which means that $k = 2$

Sample of 2, which means that $n = 2$

This is P(X = 2). So

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 2) = h(2,6,2,2) = \frac{C_{2,2}*C_{4,0}}{C_{6,2}} = 0.0667$

0.0667 = 6.67% probability that both marbles are yellow.

What is the probability of one blue and then one yellow?

Total is 100%.

Can be:

Both blue(40%)

Both yellow(6.67%)

One blue and one yellow(x%). So

40 + 6.67 + x = 100

x = 100 - 46.67

x = 53.33

53.33% probability of one blue and then one yellow.

If you are told that both selected marbles are the same color, what is the probability that both are blue?

Conditional probability.

Event A: Both same color

Event B: Both blue

Probability of both being same color:

Both blue(40%)

Both yellow(6.67%)

This means that $P(A) = 0.4 + 0.0667 = 0.4667$

Probability of both being the same color and blue:

40% probability that both are blue, which means that $P(A \cap B) = 0.4$

Desired probability:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.4}{0.4667} = 0.8571$

If you are told that both selected marbles are the same color, 0.8571 = 85.71% probability that both are blue

8 0

2 years ago

In AOPQ, o = 700 cm, p = 840 cm and q=620 cm. Find the measure of _P to the<br> nearest degree

Westkost [7]

Given:

In triangle OPQ, o = 700 cm, p = 840 cm and q=620 cm.

To find:

The measure of angle P.

Solution:

According to the Law of Cosines:

$\cos A=\dfrac{b^2+c^2-a^2}{2bc}$

Using Law of Cosines in triangle OPQ, we get

$\cos P=\dfrac{o^2+q^2-p^2}{2oq}$

$\cos P=\dfrac{(700)^2+(620)^2-(840)^2}{2(700)(620)}$

$\cos P=\dfrac{490000+384400-705600}{868000}$

$\cos P=\dfrac{168800}{868000}$

On further simplification, we get

$\cos P=0.19447$

$P=\cos^{-1}(0.19447)$

$P=78.786236$

$P\approx 79$

Therefore, the measure of angle P is 79 degrees.

8 0

3 years ago

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Estimate the sum<br> 5.89 +7 1/11

Vlad [161]

If you convert 7 1/11 to decimals it comes out to .09 there fore 5.89+7.09
Answer: 12.98

4 0

3 years ago

Read 2 more answers