Throw a dice 4 times. What is the probability 6 be up at-least one time? - Mathematics Stack Exchange most recent 30 from math.stackexchange.com 2020-01-27T07:16:28Z https://math.stackexchange.com/feeds/question/1000488 https://creativecommons.org/licenses/by-sa/4.0/rdf https://math.stackexchange.com/q/1000488 4 Throw a dice 4 times. What is the probability 6 be up at-least one time? Billie https://math.stackexchange.com/users/48863 2014-10-31T22:02:19Z 2019-11-29T23:20:39Z <p>First time I approach a probability question (:</p> <hr> <blockquote> <p>Throw a dice 4 times. What is the probability <code>6</code> be up at-least one time?</p> </blockquote> <p>Intuitively, I would say: $\frac{1}{6}\times4$.</p> <p>I would explain as: If you throw one time, probability is $\frac{1}{6}$.</p> <p>If you do it 4 times, then multiply by 4.</p> <p>According to the answers I'm wrong.</p> <p>Can you explain please? thanks in advance.</p> https://math.stackexchange.com/questions/1000488/throw-a-dice-4-times-what-is-the-probability-6-be-up-at-least-one-time/1000503#1000503 2 Answer by Mark Bennet for Throw a dice 4 times. What is the probability 6 be up at-least one time? Mark Bennet https://math.stackexchange.com/users/2906 2014-10-31T22:10:28Z 2014-10-31T22:10:28Z <p>Throw a dice six times - are you certain to get at least one six?</p> <p>The mean number of sixes in four throws is indeed $\frac 23$. But there are combinations with $2, 3 \text{ or } 4$ sixes, and these reduce the number with just one six (apply the same argument to six throws, where it is more intuitive).</p> <p>If you get no sixes in four throws, there are $5$ possibilities for each throw, and therefore $5^4=625$ possibilities with no six out of the $6^4=1296$ possibilities in total. So you get the probability by taking the $1296-625 = 671$ possibilities which must include at least one six, out of the $1296$ possibilities altogether.</p> https://math.stackexchange.com/questions/1000488/throw-a-dice-4-times-what-is-the-probability-6-be-up-at-least-one-time/1000516#1000516 1 Answer by Vladimir Vargas for Throw a dice 4 times. What is the probability 6 be up at-least one time? Vladimir Vargas https://math.stackexchange.com/users/187578 2014-10-31T22:18:48Z 2017-05-01T15:00:45Z <p>Use the binomial distribution:</p> <p>$$\mathbb P (\mbox{"obtaining x times 6 in n=4 random experiments"})=\binom{n}{x}p^xq^{n-x},$$ where $p$ is the probability of obtaining '$6$' and $q$ is $(1-p)$.</p> <p>PS: The probability of obtaining '6' at least one time is the sum of the probabilities of getting '6' one time, getting '6' two times,..., getting '6' four times.</p> https://math.stackexchange.com/questions/1000488/throw-a-dice-4-times-what-is-the-probability-6-be-up-at-least-one-time/1000655#1000655 4 Answer by Masacroso for Throw a dice 4 times. What is the probability 6 be up at-least one time? Masacroso https://math.stackexchange.com/users/173262 2014-11-01T00:28:59Z 2014-11-01T00:28:59Z <p>Expanding the comment of Vladimir Vargas about binomial distribution you can, alternatively, use the complementary probability to take zero 6's in the four throws:</p> <p>$$P(X\ge 1)=1-P(X=0)=1-\binom{4}{0}\left(\frac16\right)^0\left(1-\frac16\right)^4=1-\left(\frac56\right)^4=\frac{671}{1296}\approx52\%$$</p> https://math.stackexchange.com/questions/1000488/throw-a-dice-4-times-what-is-the-probability-6-be-up-at-least-one-time/3342714#3342714 -1 Answer by adi_226 for Throw a dice 4 times. What is the probability 6 be up at-least one time? adi_226 https://math.stackexchange.com/users/664954 2019-09-03T05:04:00Z 2019-09-03T05:04:00Z <p>Consider the complement problem, there is a 5/6 probability of not rolling a six for any given die, and since the four dice are independent, the probability of not rolling a six is (5/6)^4 = 5^4/6^4 = 625/1296. The probability of rolling at least one six is therefore 1 − 625/1296 = 671/1296 ≈ .517.</p>