Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Find the probability that a person flipping a balanced coin requires four tosses to get a head.

Short Answer

Expert verified
The probability that a person flipping a balanced coin requires four tosses to get their first head is \(1/16\).

Step by step solution

01

Determine the probability of each toss

Since the coin is balanced, the probabilities for getting a head (H) or a tail (T) in each toss are equal and independent of each other. Therefore, the probability of getting a head in any toss is P(H) = 1/2, and the probability of getting a tail in any toss is P(T) = 1/2.
02

Calculate the probability of the desired sequence

The desired sequence in this case is TTT H, which means getting three tails and then a head. Since the events are independent, the probability of this sequence is the product of the probabilities of each event: P(TTT H) = P(T) * P(T) * P(T) * P(H).
03

Substitute the probabilities of each event

Now we will substitute the probabilities of each event in the formula calculated in Step 2. P(T) = 1/2 and P(H) = 1/2, so the probability of the sequence TTT H is: P(TTT H) = (1/2) * (1/2) * (1/2) * (1/2).
04

Simplify the expression and find the probability

Simplify the expression by multiplying the fractions together: P(TTT H) = (\(1/2\))^4 = 1/16. The probability that a person flipping a balanced coin requires four tosses to get their first head is 1/16.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

If \(Z\) is a standard normal variable, use the table of standard normal probabilities to find: (a) \(\operatorname{Pr}(z<0)\) (b) \(\operatorname{Pr}(-12.54)\)

A research worker wishes to estimate the mean of a population using a sample large enough that the probability will be \(.95\) that the sample mean will not differ from the population mean by more than 25 percent of the standard deviation. How large a sample should he take?

Two individuals agree to meet at a certain spot sometime between 5:00 and 6:00 P.M. They will each wait 10 minutes starting from when they arrive. If the other person does not show up, they will leave. Assume the arrival times of the two individuals are Independent and uniformly distributed over the hour- long interval, find the probability that the two will actually meet.

In one income group, \(45 \%\) of a random sample of people express approval of a product. In another income group, \(55 \%\) of a random sample of people express approval. The standard errors for these percentages are \(.04\) and \(.03\) respectively. Test at the \(10 \%\) level of significance the hypothesis that the percentage of people in the second income group expressing approval of the product exceeds that for the first income group.

Out of a group of 10,000 degree candidates of The University of North Carolina at Chapel Hill, a random sample of 400 showed that 20 per cent of the students have an earning potential exceeding \(\$ 30,000\) annually. Establish a \(.95\) confidence- interval estimate of the number of students with a \(\$ 30,000\) plus earning potential.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free