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Dennis_Churaev [7]
3 years ago
8

A psychology experiment on memory was conducted which required participants to recall anywhere from 1 to 10 pieces of informatio

n. Based on many results, the (partial) probability distribution below was determined for the discrete random variable (X = number of pieces of information remembered (during a fixed time period)). What is the missing probability P(X=7)? Your answer should include the second decimal place.
Mathematics
1 answer:
blagie [28]3 years ago
8 0

Answer:

The missing probability is, P (X = 7) = 0.24.

Step-by-step explanation:

The complete question is:

A psychology experiment on memory was conducted which required participants to recall anywhere from 1 to 10 pieces of information. Based on many results, the (partial) probability distribution below was determined for the discrete random variable (X = number of pieces of information remembered (during a fixed time period)).

What is the missing probability P(X=7)? Your answer should include the second decimal place.

X = # information | probability:

1 | 0.0

2 | 0.02

3 | 0.04

4 | 0.07

5 | 0.15

6 | 0.18

7 | ?

8 | 0.14

9 | 0.11

10 | 0.05

Solution:

The sum of the probabilities of all events of an experiment is always 1.

\sum\limits^{n}_{i=1}{P(X=x_{i})}=1

Use the above theorem to compute the missing probability.

\sum\limits^{n}_{i=1}{P(X=i)}=1

P(X=1)+P(X=2)+P(X=3)+...+P(X=10)=1\\0.00+0.02+0.04+0.07+0.15+0.18+a+0.14+0.11+0.05=1\\0.76+a=1\\a=1-0.76\\a=0.24

Thus, the missing probability is, P (X = 7) = 0.24.

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A line is a simple geometric shape that extends in both the directions, but a line segment has two defined endpoints. Both the figures are also different from a ray, as a ray has only one endpoint and extend infinitely in one direction.

Step-by-step explanation:

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Assume that 12 people, including the husband and wife pair, apply for 4 sales positions. People are hired at random.
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brainly.com/question/5218999


The formula C(n, r)= \frac{n!}{r!(n-r)!}, where r! is 1*2*3*...r

is the formula which gives us the total number of ways of forming groups of r objects, out of n objects.

for example, given 10 objects, there are C(10,6) ways of forming groups of 6, out of the 10 objects.

-----------------------------------------------------------------------------------------------


Selecting 4 people out of 12 can be done in :

\displaystyle{C(12, 4)= \frac{12!}{4!8!}= \frac{12\cdot11\cdot10\cdot9\cdot8!}{4!8!}= \frac{12\cdot11\cdot10\cdot9}{4!}=11\cdot5\cdot9= 495       many ways.


All the possible groups of 4 people, where the husband and wife are included, can be done in C(10, 2) many ways, since we only calculate the possible choices of 2 out of 10 people, to complete the groups of 4.


\displaystyle{ C(10, 2)= \frac{10!}{2!8!}= \frac{10\cdot9}{2}=45


Thus, the 

probability that both the husband and wife are hired is 45/495=0.09


Part 2)

The probability that one is hired and the other is not = 

P(husband hired, wife not hired) + P(wife hired, husband not hired)

these 2 are clearly equal, so it is enough to calculate one.


Consider the case : husband hired, wife not hired.

assuming the husband is hired, we have to calculate the possible groups of 3 that can be formed from 11-1 (the wife)=10 people.

this is 

\displaystyle{ C(10, 3)= \frac{10!}{3!7!}= \frac{10 \cdot9 \cdot8}{3\cdot2}=10\cdot3\cdot4=120


thus, 


P(husband hired, wife not hired)=120/495=0.24


thus, 

The probability that one is hired and the other is not = 

P(husband hired, wife not hired) + P(wife hired, husband not hired) =

0.24+0.24=0.48



Answer:


A) 0.09


B) 0.48

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