What is the meaning of two unbiased dice?
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friends at today's question is two unbiased dice are thrown find the probability that the total of the number on the dice is it it is given to us that two dice are thrown that unbiased that is the probability of getting each number will be 1 by 6 or we can see equal then find the probability that the total of the numbers on the dice is it we have to find the probability that the sum of the numbers will be it so it is solution part the number of elements in the samples were we can see total out possible outcomes are total possible outcomes have 6 x 6 that is 3636 are the total possible as in the first time we can have the numbers from 1 to 6 1 2 3 4 5 6 and in the second I to we can have the numbers from 1 to 6 so Show
Unnati ordered pair it will be 6 x 6 that is 36 will have 36 element now what is the condition they went to us that the number of the sum of the number will be it so the sum for the sum of integers to be a twihard having the possibilities that oneplus 7 we can have these numbers oneplus 7 2 + 6 oneplus 7 is not possible as on diets on dice with only one or only the numbers from 1 to 6 that is so we will have to neglect this case we cannot have this so we will have to + 63 + 54 + 45 + 3 and 6 plus 2 only 10 cases we are having to on the first S6 on the second 3 on the first and the second 4 on the first and follow in the second on the first three on the second and six on the first two on the second this case we are having to how to find the probability to find the probability we have to divide number of favourable outcomes favourable outcomes by we have to divide number of favourable outcomes by total outcomes this is the way to find probability of this this this this and this so we're having 5 favourable outcomes to have the value of probabilities 5 / total outcomes at 36 25 36 so this is the probability that we have to calculate Two unbiased dice are thrown. If the probability that the sum of the numbers obtained on the two dice is neither a multiple of 2 nor a multiple of 3 is \[\dfrac{1}{p}\], find the value of p.Answer Verified Hint: Write the total possible outcome that might occur in throwing two unbiased dice. From that formulate the pairs which will give the sum 2, 3, 4, 6, 8, 9, 10, 12 which are the multiples of either 2 or 3. Find the probability of its occurrence which will be P (E). Then find P (not E) i.e. we need to find the pair that don’t sum up to some multiple of 2 or 3. Complete step-by-step answer: Therefore, the possible outcome that can be obtained from two dice are \[6\times 6=36\]. \[\therefore n={{6}^{2}}=36\]. The possible outcomes are listed below. \[\left\{ \begin{align} & \left( 2,1 \right),\left( 2,2 \right),\left( 2,3 \right),\left( 2,4 \right),\left( 2,5 \right),\left( 2,6 \right), \\ & \left( 3,1 \right),\left( 3,2 \right),\left( 3,3 \right),\left( 3,4 \right),\left( 3,5 \right),\left( 3,6 \right), \\ & \left( 4,1 \right),\left( 4,2 \right),\left( 4,3 \right),\left( 4,4 \right),\left( 4,5 \right),\left( 4,6 \right), \\ & \left( 5,1 \right),\left( 5,2 \right),\left( 5,3 \right),\left( 5,4 \right),\left( 5,5 \right),\left( 5,6 \right), \\ & \left( 6,1 \right),\left( 6,2 \right),\left( 6,3 \right),\left( 6,4 \right),\left( 6,5 \right),\left( 6,6 \right) \\ \end{align} \right\}\] Let E be the event that the sum of the numbers obtained on the dice is neither a multiple of 2 nor a multiple of 3. Let E’ be the event that sum of numbers obtained on the dice is either multiple of 2 or 3, where 2, 4, 6, 8, 10 and 12 are multiples of 2, and 3, 6, 9, 12 are multiples of 3. Here 12 is taken as the value of highest sum, as it is the sum of numbers obtained by adding the highest number in each dice. 6 is the highest number in both the dice. Therefore, the sum = 6 + 6 = 12 is the highest sum for two unbiased dice. Therefore, the sample space of the event E’ can be found from the total possible outcomes. \[\left\{ \begin{align} & \left( 3,1 \right),\left( 3,3 \right),\left( 3,5 \right),\left( 3,6 \right),\left( 4,2 \right),\left( 4,4 \right),\left( 4,5 \right),\left( 4,6 \right), \\ & \left( 5,1 \right),\left( 5,3 \right),\left( 5,4 \right),\left( 5,5 \right),\left( 6,2 \right),\left( 6,3 \right),\left( 6,4 \right),\left( 6,6 \right) \\ \end{align} \right\}\] So there are 24 favorable possibilities out of 36 total possibilities. \[\therefore n\left( E' \right)=24\] Total possibilities, \[n\left( s \right)=36\]. Therefore, probability of an event E is given by P (E). P (E) = number of simple events within THE / total number of possible outcomes. \[P\left( E' \right)=\dfrac{n\left( E' \right)}{n\left( s \right)}=\dfrac{24}{36}\] We know that \[P\left( E \right)=1-P\left( E' \right)\], where \[P\left( E' \right)\]can also be said as P (not E). \[\therefore P\left( E
\right)=1-\dfrac{24}{36}\] \[\therefore P\left( E \right)=\dfrac{1}{3}=\dfrac{1}{p}\] From the question it was told that the probability is \[\dfrac{1}{p}\]. We got \[P\left( E \right)=\dfrac{1}{3}\]. Equate, \[\dfrac{1}{p}=\dfrac{1}{3}\]. Cross multiplying, we get p = 3. Hence we got the value of p = 3. Note: We found the number of pairs which give a sum that are multiples of either 2 or 3. What is the meaning of unbiased in probability?Unbiased coin means that the probability of heads is the same as the probability of tails, each being 1/2(equal probability of selection),. A coin that has two different sides for two different results,irrespctive of how many trials you do.
What is difference between biased and unbiased?In a biased sample, one or more parts of the population are favored over others, whereas in an unbiased sample, each member of the population has an equal chance of being selected.
When 2 unbiased dice are rolled what is the probability that the sum on the top of the faces is greater than 10?Detailed Solution
∴ The probability that the sum on the top of the faces is greater than 10 is 1/12.
What is the probability of rolling two unbiased dice rolled together?It is said that two unbiased dice are drawn. The total number of possible outcomes from one dice is 6. Therefore, the possible outcome that can be obtained from two dice are \[6\times 6=36\].
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