/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 41 If a die is rolled 4 times, what... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If a die is rolled 4 times, what is the probability that 6 comes up at least once?

Short Answer

Expert verified
The probability of getting at least one 6 in four rolls of a die is \(\frac{671}{1296}\).

Step by step solution

01

Determine the probability of not getting a 6 in one roll

Since a fair die has 6 faces, the possibility of not getting a 6 in one roll is the probability of getting any of the other 5 faces. Therefore, the probability of not getting a 6 in one roll is 5/6.
02

Determine the probability of not getting a 6 in all four rolls

The probability of not getting a 6 in all four rolls can be found by multiplying the probability of not getting a 6 in one roll by itself four times. Mathematically, this can be expressed as \((\frac{5}{6})^4\).
03

Calculate the probability of not getting a 6 in all four rolls

Using the formula from step 2, we calculate the probability of not getting a 6 in all four rolls: \((\frac{5}{6})^4 = \frac{625}{1296}\).
04

Calculate the probability of getting at least one 6 in the four rolls

Now, to find the probability of getting at least one 6 in the four rolls, we need to subtract the probability of not getting a 6 in all four rolls from 1: \(1 - \frac{625}{1296} = \frac{671}{1296}\).
05

Simplify the probability

The probability of getting at least one 6 in the four rolls is \(\frac{671}{1296}\). This fraction is already in its simplest form, so the final answer is \(\boxed{\frac{671}{1296}}\).

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 91Ó°ÊÓ!

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

The chess clubs of two schools consist of, respectively, 8 and 9 players. Four members from each club are randomly chosen to participate in a contest between the two schools. The chosen players from one team are then randomly paired with those from the other team, and each pairing plays a game of chess. Suppose that Rebecca and her sister Elise are on the chess clubs at different schools. What is the probability that (a) Rebecca and Elise will be paired? (b) Rebecca and Elise will be chosen to represent their schools but will not play cach other? (c) either Rebecca or Elise will be chosen to represent her school?

How many people have to be in a room in order that the probability that at least two of them celebrate their birthday in the same month is at least \(\frac{1}{2} ?\) Assume that all possible monthly outcomes are equally likely.

Poker dice is played by simultaneously rolling 5 dice. Show that (a) \(P\\{\text { no two alike }\\}=.0926\) (b) \(P\\{\text { one pair }\\}=.4630\) (c) \(P\\{\text { two pair }\\}=.2315\) (d) \(P\\{\text { three alike }\\}=.1543\) (e) \(P\\{\text { full house }\\}=.0386\) (f) \(P\\{\text { four alike }\\}=.0193\) (g) \(P\\{\text { five alike }\\}=.0008\)

A forest contains 20 elk, of which 5 are captured, tagged, and then released. A certain time later, 4 of the 20 elk are captured. What is the probability that 2 of these 4 have been tagged? What assumptions are you making?

A hospital administrator codes incoming patients suffering gunshot wounds according to whether they have insurance (coding 1 if they do and 0 if they do not) and according to their condition, which is rated as good \((g),\) fair \((f),\) or serious (s). Consider an experiment that consists of the coding of such a patient. (a) Give the sample space of this experiment. (b) Let \(A\) be the event that the patient is in serious condition. Specify the outcomes in \(A\) (c) Let \(B\) be the event that the patient is uninsured. Specify the outcomes in \(B\). (d) Give all the outcomes in the event \(B^{c} \cup A\).

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.