What Is the Probability of Rolling a 3 on a Four-Sided Die?

In summary, the problem involves finding the probability of event A, the event that a 3 is observed on the first trial, and event B, the event that at least two trials are required to observe a 3, when rolling a four-sided die. The sample space is S={1,2,3,4}, and the probabilities of event A and B are 1/4 and 3/16, respectively. The probability of event A U B, the event that either event A or B (or both) occurs, is 7/16.
  • #1
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Statistic Help!THANKS!

Consider the trial on which a 3 is first observed in successive rolls of a four-sided die. Let A be the event that 3 is observed on the first trial. Let B be the event that at least two trials are required to observed a 3. assuming that each side has probability 1/4. find (a)P(A) (b)P(B),and (c) P(A U B)I think the S={1,2,3,4} and event A={3}, but i was confuse with the event B.

(a) P(A)=1-P(A')=1-3/4=1/4

But since i don't know event B, i can't do the rest of problem.
 
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  • #2


Hi there,

I'm glad to help you with this problem! Let's break it down step by step.

First, let's define the experiment and sample space. The experiment is rolling a four-sided die, and the sample space, as you correctly identified, is S={1,2,3,4}.

Next, let's define the events. Event A is the event that a 3 is observed on the first trial. This means that the first roll must be a 3, and the other three rolls can be any number. So, the probability of event A is simply the probability of rolling a 3 on the first roll, which is 1/4.

Event B is the event that at least two trials are required to observe a 3. This means that the first roll must not be a 3, and the second, third, or fourth roll must be a 3. So, the probability of event B is the probability of not rolling a 3 on the first roll (3/4) multiplied by the probability of rolling a 3 on the second, third, or fourth roll (1/4), which is (3/4)*(1/4)=3/16.

Finally, event A U B (A union B) is the event that either event A or event B (or both) occurs. In other words, it is the event that a 3 is observed on the first trial OR at least two trials are required to observe a 3. To calculate the probability of this event, we can simply add the probabilities of event A and event B, since they are mutually exclusive (they cannot occur at the same time). So, P(A U B)=P(A)+P(B)=(1/4)+(3/16)=7/16.

I hope this helps clarify the problem for you. Let me know if you have any further questions. Good luck with your studies!
 

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