Probability question (Please help)

[Moonlightbeam]

I don't know how to solve this multiple choice question. Is there supposed to be an equation I am supposed to use.

In the expression 12C1(0.35)¹¹(0.65), the probability of success is:

• 7/20
• 35%
• 13/20
• 11/12

 

[pka]

I don't know how to solve this multiple choice question. Is there supposed to be an equation I am supposed to use.
In the expression 12C1(0.35)¹¹(0.65), the probability of success is:

• 7/20
• 35%
• 13/20
• 11/12

• Have a look Here.

 

[Steven G]

They do not tell you what 12C1(0.35)¹¹(0.65) means. The answer is either .35 or .65.
In my opinion, pka misread the problem.

 

[Moonlightbeam]

They do not tell you what 12C1(0.35)¹¹(0.65) means. The answer is either .35 or .65.
In my opinion, pka misread the problem.

Why would the answer be either .35 or .65. I confused on what the 12C1 is supposed to represent in the question

 

[pka]

They do not tell you what 12C1(0.35)¹¹(0.65) means. The answer is either .35 or .65.
In my opinion, pka misread the problem.

What is there to missread as posted is it not \(\dbinom{12}{1}(0.35)^{11}(0.65)~?\)
How do you read what is posted?

 

[Steven G]

As far as the choices go, it is either 7/20 or 13/20

 

[Steven G]

What is there to missread as posted is it not \(\dbinom{12}{1}(0.35)^{11}(0.65)~?\)
How do you read what is posted?

The probability of success is .35 and the probability of failure is .65 OR the probability of success is .65 and the probability of failure is .35
That is how I read what is posted.

 

[Steven G]

Why would the answer be either .35 or .65. I confused on what the 12C1 is supposed to represent in the question

The expression given is either p(having exactly 1 success out of 12) or p(having exactly 11 success out of 12).
But they did not ask for either of those probabilities!

 

[Cubist]

That expression looks very much like a binomial distribution's probability mass function (Wikipedia link). My guess is that the question ought to be interpreted like this (given the possible responses)...

The probability of one success in 12 independent Bernoulli trials is given by the expression 12C1 * (0.35)^11 * (0.65). What is the probability of success in each individual Bernoulli trial ?

 

[pka]

The probability of success is .35 and the probability of failure is .65 OR the probability of success is .65 and the probability of failure is .35 That is how I read what is posted.

The expression given is either p(having exactly 1 success out of 12) or p(having exactly 11 success out of 12). But they did not ask for either of those probabilities!

BUT that is not what was posted.
\(\dbinom{12}{1}(0.35)^{11}(0.65)=0.000075308342750976562500000\)
\(\dbinom{12}{1}(0.35)(0.65)^{11}=0.036753289330836914062500\)

 

[Steven G]

BUT that is not what was posted.
\(\dbinom{12}{1}(0.35)^{11}(0.65)=0.000075308342750976562500000\)
\(\dbinom{12}{1}(0.35)(0.65)^{11}=0.036753289330836914062500\)

The 2nd one was not posted.

Look, it is not hard to see what I am saying.
Suppose you have 12 light bulbs in a box. The probability that anyone is good (let's say that is a success) is .35. Yes, the p(that exactly 11 are good out of the 12) = 0.000075308342750976562500000.

Here is where we disagree. The problem asks for the probability of success NOT the prob that exactly 11 of the 12 bulbs are good!

Cubist agrees with my interpretation.