Elena Marquez

2022-10-02

Why is Binomial Probability used here?

A test consists of 10 multiple choice questions with five choices for each question. As an experiment, you GUESS on each and every answer without even reading the questions.What is the probability of getting exactly 6 questions correct on this test?

The answer says:

$$P={\textstyle (}\genfrac{}{}{0ex}{}{10}{6}{\textstyle )}(0.2{)}^{6}(0.8{)}^{4}$$

I was thinking though:

$$P(A\text{and}B)=P(A)\cdot P(B)$$

Probability of getting 6 right and 4 wrong would be just:

$$(0.2{)}^{6}(0.8{)}^{4}$$

Why is the binomial coefficient used there?

A test consists of 10 multiple choice questions with five choices for each question. As an experiment, you GUESS on each and every answer without even reading the questions.What is the probability of getting exactly 6 questions correct on this test?

The answer says:

$$P={\textstyle (}\genfrac{}{}{0ex}{}{10}{6}{\textstyle )}(0.2{)}^{6}(0.8{)}^{4}$$

I was thinking though:

$$P(A\text{and}B)=P(A)\cdot P(B)$$

Probability of getting 6 right and 4 wrong would be just:

$$(0.2{)}^{6}(0.8{)}^{4}$$

Why is the binomial coefficient used there?

tona6v

Beginner2022-10-03Added 6 answers

Step 1

The short answer is that order matters.

The long answer begins with an example. Suppose you had only 3 questions, and each question was True/False, so you have equal probabilities of answering any given question correctly by random guessing. Then you can easily enumerate the possible outcomes of whether or not you get Questions 1, 2, or 3 correct. Let the ordered triplet $({q}_{1},{q}_{2},{q}_{3})$ represent the outcomes of questions 1,2,3 in that order, where ${q}_{i}\in \{R,W\}$ where R indicates a right answer, and W indicates a wrong answer. Then your outcomes are

$$\begin{array}{r}(R,R,R)\\ (R,R,W)\\ (R,W,R)\\ (R,W,W)\\ (W,R,R)\\ (W,R,W)\\ (W,W,R)\\ (W,W,W)\end{array}$$

and now it is obvious that there are 8 elementary outcomes. If I asked you what the probability is of getting exactly 2 correct answers out of three, you would say 3/8. But your method of reasoning for the original question would have given $(1/2{)}^{3}=1/8$, which is clearly not the case.

So why is your method not correct? Because there are, in general, $(}\genfrac{}{}{0ex}{}{n}{k}{\textstyle )$ arrangements of k correct answers out of n questions, and each distinct ordering of correct/incorrect answers is a distinct elementary outcome.

The short answer is that order matters.

The long answer begins with an example. Suppose you had only 3 questions, and each question was True/False, so you have equal probabilities of answering any given question correctly by random guessing. Then you can easily enumerate the possible outcomes of whether or not you get Questions 1, 2, or 3 correct. Let the ordered triplet $({q}_{1},{q}_{2},{q}_{3})$ represent the outcomes of questions 1,2,3 in that order, where ${q}_{i}\in \{R,W\}$ where R indicates a right answer, and W indicates a wrong answer. Then your outcomes are

$$\begin{array}{r}(R,R,R)\\ (R,R,W)\\ (R,W,R)\\ (R,W,W)\\ (W,R,R)\\ (W,R,W)\\ (W,W,R)\\ (W,W,W)\end{array}$$

and now it is obvious that there are 8 elementary outcomes. If I asked you what the probability is of getting exactly 2 correct answers out of three, you would say 3/8. But your method of reasoning for the original question would have given $(1/2{)}^{3}=1/8$, which is clearly not the case.

So why is your method not correct? Because there are, in general, $(}\genfrac{}{}{0ex}{}{n}{k}{\textstyle )$ arrangements of k correct answers out of n questions, and each distinct ordering of correct/incorrect answers is a distinct elementary outcome.

overrated3245w

Beginner2022-10-04Added 3 answers

Step 1

We know that the total probability for any event happening for all cases must be 1. If the problem was solved the way you presented, would the total probabilities add up to 1? Clearly not.

Let's look at each individual question, which by itself has probabilities that add up to one. For a particular question, getting it right is a 0.2 chance and getting it wrong is a 0.8 chance, a total of 1. If we go question by question, we take the probability to a power, so let's do that with the total probability:

$$\begin{array}{rl}(0.8+0.2)& =0.8+0.2\\ (0.8+0.2{)}^{2}& ={0.8}^{2}+2(0.8)(0.2)+{0.2}^{2}\\ (0.8+0.2{)}^{3}& ={0.8}^{3}+3(0.8{)}^{2}(0.2)+3(0.8)(0.2{)}^{2}+{0.2}^{3}\\ (0.8+0.2{)}^{4}& ={0.8}^{4}+4(0.8{)}^{3}(0.2)+6(0.8{)}^{2}(0.2{)}^{2}+4(0.8)(0.2{)}^{3}+{0.2}^{4}\end{array}$$

Step 2

And so on and so forth. This is a direct application of the binomial theorem:

$$(x+y{)}^{n}=\sum _{k=0}^{n}{\textstyle (}\genfrac{}{}{0ex}{}{n}{k}{\textstyle )}{x}^{n-k}{y}^{k}$$

for $x=0.8$ and $y=0.2$

We know that the total probability for any event happening for all cases must be 1. If the problem was solved the way you presented, would the total probabilities add up to 1? Clearly not.

Let's look at each individual question, which by itself has probabilities that add up to one. For a particular question, getting it right is a 0.2 chance and getting it wrong is a 0.8 chance, a total of 1. If we go question by question, we take the probability to a power, so let's do that with the total probability:

$$\begin{array}{rl}(0.8+0.2)& =0.8+0.2\\ (0.8+0.2{)}^{2}& ={0.8}^{2}+2(0.8)(0.2)+{0.2}^{2}\\ (0.8+0.2{)}^{3}& ={0.8}^{3}+3(0.8{)}^{2}(0.2)+3(0.8)(0.2{)}^{2}+{0.2}^{3}\\ (0.8+0.2{)}^{4}& ={0.8}^{4}+4(0.8{)}^{3}(0.2)+6(0.8{)}^{2}(0.2{)}^{2}+4(0.8)(0.2{)}^{3}+{0.2}^{4}\end{array}$$

Step 2

And so on and so forth. This is a direct application of the binomial theorem:

$$(x+y{)}^{n}=\sum _{k=0}^{n}{\textstyle (}\genfrac{}{}{0ex}{}{n}{k}{\textstyle )}{x}^{n-k}{y}^{k}$$

for $x=0.8$ and $y=0.2$

Assume that when adults with smartphones are randomly selected, 54% use them in meetings or classes (based on data from an LG Smartphone survey). If 8 adult smartphone users are randomly selected, find the probability that exactly 6 of them use their smartphones in meetings or classes?

Write formula for the sequence of -4, 0, 8, 20, 36, 56, 80, where the order of f(x) is 0, 1, 2, 3, 4, 5, 6 respectively

A binomial probability experiment is conducted with the given parameters. Compute the probability of x successes in the n independent trials of the experiment.n=20,

p=0.7,

x=19

P(19)=

In binomial probability distribution, the dependents of standard deviations must includes.

a) all of above.

b) probability of q.

c) probability of p.

d) trials.The probability that a man will be alive in 25 years is 3/5, and the probability that his wifewill be alive in 25 years is 2/3

Determine the probability that both will be aliveHow many different ways can you make change for a quarter??

(Different arrangements of the same coins are not counted separately.)One hundred people line up to board an airplane that can accommodate 100 passengers. Each has a boarding pass with assigned seat. However, the first passenger to board has misplaced his boarding pass and is assigned a seat at random. After that, each person takes the assigned seat. What is the probability that the last person to board gets his assigned seat unoccupied?

A) 1

B) 0.33

C) 0.6

D) 0.5The value of $(243{)}^{-\frac{2}{5}}$ is _______.

A)9

B)$\frac{1}{9}$

C)$\frac{1}{3}$

D)01 octopus has 8 legs. How many legs does 3 octopuses have?

A) 16

B 24

C) 32

D) 14From a pack of 52 cards, two cards are drawn in succession one by one without replacement. The probability that both are aces is...

A pack of cards contains $4$ aces, $4$ kings, $4$ queens and $4$ jacks. Two cards are drawn at random. The probability that at least one of these is an ace is A$\frac{9}{20}$ B$\frac{3}{16}$ C$\frac{1}{6}$ D$\frac{1}{9}$

You spin a spinner that has 8 equal-sized sections numbered 1 to 8. Find the theoretical probability of landing on the given section(s) of the spinner. (i) section 1 (ii) odd-numbered section (iii) a section whose number is a power of 2. [4 MARKS]

If A and B are two independent events such that $P(A)>0.5,P(B)>0.5,P(A\cap \overline{B})=\frac{3}{25}P(\overline{A}\cap B)=\frac{8}{25}$, then the value of $P(A\cap B)$ is

A) $\frac{12}{25}$

B) $\frac{14}{25}$

C) $\frac{18}{25}$

D) $\frac{24}{25}$The unit of plane angle is radian, hence its dimensions are

A) $[{M}^{0}{L}^{0}{T}^{0}]$

B) $[{M}^{1}{L}^{-1}{T}^{0}]$

C) $[{M}^{0}{L}^{1}{T}^{-1}]$

D) $[{M}^{1}{L}^{0}{T}^{-1}]$Clinical trial tests a method designed to increase the probability of conceiving a girl. In the study, 340 babies were born, and 289 of them were girls. Use the sample data to construct a 99% confidence interval estimate of the percentage of girls born?