How to simplify this

Suppose box A contains 4 red and 5 blue poker chips and box B contains 6 red and 3 blue poker chips. Then a poker chip is chosen

sergejj [24]

Answer:

0.5172 = 51.72% probability a blue poker chip was transferred from box A to box B, given that the coin just chosen from box B is red

Step-by-step explanation:

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.

In this question:

Event A: Coin chosen from box B is red.

Event B: Blue poker chip transferred.

Probability of choosing a red coin:

7/10 of 4/9(red coin from box A)

6/10 of 5/9(blue coin from box A). So

$P(A) = \frac{7}{10}*\frac{4}{9} + \frac{6}{10}*\frac{5}{9} = \frac{28 + 30}{90} = 0.6444$

Blue chip transferred, red coin chosen:

6/10 of 5/9. So

$P(A \cap B) = \frac{6}{10}*\frac{5}{9} = \frac{30}{90} = 0.3333$

What is the probability a blue poker chip was transferred from box A to box B, given that the coin just chosen from box B is red?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.3333}{0.6444} = 0.5172$

0.5172 = 51.72% probability a blue poker chip was transferred from box A to box B, given that the coin just chosen from box B is red

5 0

2 years ago

The design of a microchip has the scale 40:1. The length of the design is 18cm, find the actual length of the micro chip?

zimovet [89]

Answer:

0.45 cm

Step-by-step explanation:

Actual length of the micro chip

$= \frac{1}{40} \times 18 \\ \\ =0.45 \: cm$

4 0

3 years ago

Is 85/95 greater than 64/76

elena-14-01-66 [18.8K]

The answer is yes I believe

5 0

3 years ago

How would I do this problem?

kenny6666 [7]

If it's a geometric sequence then:

$a_1=27;\ a_2=27\\\\r=\dfrac{a_2}{a_1}\to r=\dfrac{27}{36}=\dfrac{3}{4}=0.75\\\\a_{n+1}=a_nr\\\\a_3=27\cdot0.75=20.25\ CORRECT$

We calculate the fourth and fifth term of the sequence:

$a_4=a_3r\to a_4=20.25\cdot0.75=15.1875\\\\a_5=a_4r\to a_5=15.1875\cdot0.75=11.390625$

Answer:

In year 4 15.1875 animals.

In year 5 11.390625 animals.

7 0

3 years ago

At a financial institution, a fraud detection system identifies suspicious transactions and sends them to a specialist for revie

labwork [276]

Answer:

a. E(X) = 54.4

b. E(X) = 2.5

c. P(Y=2) = .0116

Step-by-step explanation:

a.

E(X) = np = .40 probability * 136 trials = 54.4 blocked transmissions

To get the expected value, we simply multiply probability times number of trials. You can look at it in simple terms by thinking if there's a 50% chance of flipping heads and you flip a coin twice, in an ideal world you will have .5*2 = 1 head.

b.

i. Let X represent the number of suspicious transmissions reviewed until finding the first blocked one. We will use a geometric distribution to model the "first" transmission. Whenever we're looking for the "first" time something happens, we use geometric.

ii. E(X) = 1/p , according to the geometric model.

= 1/.4 = 2.5.

We expect that the specialist will review 2.5 suspicious transactions <em>on average </em>before finding the first transmission that will be blocked.

c.

i. Let Y represent the exact number of blocked transmissions out of 10. We will use a binomial distribution to model the "fixed" number of transmissions. Whenever we're looking for a "fixed" number of times something happens, we use binomial.

ii. P(Y=k) = (n choose k)(p^k)(q^n-k)

P(Y=2) = (¹⁰₂)(.4^2)(.6^10-2)

= 45 (.4^2)(.6^10-2) = .0016

As for calculator notation, the n choose k can be accessed on a TI-84 via MATH -> PRB -> nCr. It looks like 10 nCr 2 on the display.

Hence the probability that two transactions out of ten will be blocked is .0016 by the binomial model.

5 0

3 years ago