Alice spins the spinner 3 times. What is the probability that she spins an A, then a B, and then a C? http://static.k12.com/calms_media/media/332000_332500/332448/2/580a1a69ea83d3dd36df179e9d8ab7a5003d26d9/MS_ALG_A2_04_quiz_Q2_question.gif A. 1/3 B. 1/9 C. 1/27 D. 1/12
@iGreen
@sonic400539 Very smart way of tagging people one at a time without leaving a trace! Anyway, what is the probability of landing on A for the first spin?
Lol^
1/3
The probability of spinning A, B, or C are the same, so when we find one, we can multiply it by itself 2 times.
Yep, you got it.
Now multiply it by itself 2 times.
xD Yeah, I always clear them so I don't look desperate.
Good, what is the probability of landing on B in a subsequent spin?
2/3?
B is still one third of the whole circle, so there is not a better probability than A, do you agree?
Yes, I agree. It makes sense.
What about C, again in a subsequent spin?
1/3
All have an equal chance, correct?
Yes, each one has equal chance. Now, the experiment of getting A, then B, then C is called a multi-step experiment. The probability is the product of the three individual probabilities IF the steps are independent of each other.
Here, independence means that the result of one spin does NOT affect the result of subsequent spins, which is the case here, so the multiplication rule mentioned above applies.
@sonic400539 ok so far?
I'm following it. (So far.)
So how would you find the answer to the problem? If you have doubts, reread what I just posted to make sure you understand it. It is the key to your answering ALL similar problems.
I believe that it is B because 1/3 is A, B, and C and when added you have a total probability of 1/9.
Wait, the answer is C. 1/27. The chances are correct. Thanks for the help! Question: Closed
Exactly: Good work! \(\color{blue}{\Large\checkmark}{Way~to~go!}\)
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