It follows that the **number** **of** **possible** combinations is equal to 6 x 6 = 36 as each of the two **dice** used in craps has six sides. Each **throw** **of** **the** two **dice** will result in one of these eleven **numbers** coming out. Some **numbers**, however, are rolled more frequently as the **number** **of** combinations that add up to them is greater. Total **number** **of** ways = 6 x 6 x ... to n times = 6nTotal **number** **of** ways to show only even **numbers**= 3 x 3 x ... to n times = 3n.∴the required **number** **of** ways = 6n - 3n. The **number** **of** **possible** **outcomes** **in** **a** **throw** **of** n **ordinary** **dice** **in** which at least one of the **dice** shown an odd **number** isa)6n-1b)3n-1c)6n-3nd)none of theseCorrect answer is option 'B'.

**The number** **of possible** **outcomes** **in a throw** of n **ordinary** **dice** in which at least one of the **dice** shows on odd **number** is Moderate A 6n−1 B 3n−1 C 6n−3n D nP6− nP3 Solution **The number** of **outcomes** of n **dice** =6×6×......n times=6n **The number** of **outcomes** when there is no odd **number** (all even **number**) on any die =3×3×3×......n times=3n. Web.

**The number** **of possible** **outcomes** **in a throw** of n **ordinary** **dice** in which at least one of the **dice** shows on odd **number** is Moderate A 6n−1 B 3n−1 C 6n−3n D nP6− nP3 Solution **The number** of **outcomes** of n **dice** =6×6×......n times=6n **The number** of **outcomes** when there is no odd **number** (all even **number**) on any die =3×3×3×......n times=3n. Web.

**The** Question and answers have been prepared according to the CAT exam syllabus. Information about The **number** **of** distinct **outcomes** **possible** with **a** **throw** **of** 3 indistinguishable **dice** isa)216b)36c)56d)72Correct answer is option 'C'.

Web. Total **number** of ways = 6 × 6 × ..... to n times = 6n.Total **number** of ways to show only even **number** = 3 × 3 × ..... to n times = 3n.∴ required **number** of ways = 6n – 3n. **The number** **of possible** **outcomes** **in a throw** of n **ordinary** **dice** in which at least one of the **dice** shows an odd **number** isa)6n – 1b)3n – 1c)6n – 3nd)None of .... Web.

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Click here👆to get an answer to your question ️ **The number** **of possible** **outcomes** **in a throw** of n **ordinary** **dice** in which at least one of the **dice** shows an odd **number** is. Jan 06, 2020 · **The number of possible outcomes in a throw** of `n` **ordinary** **dice** in which at least one of the die shows and odd **number** is a. `6^n-1` b. `3^n-1` c. `6^n-3^n` d. none of these A. `6^(n)-1` B. `3^(n)-1`.

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(ii) if a die is thrown, there are two **possible** **outcomes**- **an** odd **number** or an even **number** .therefore the probability of getting an odd **number** is 1/2 total **outcomes** that can occur are 1, 2, 3, 4, 5, 6 **number** **of** **possible** **outcomes** **of** **a** **dice** = 6 **numbers** which are odd = 1, 3, 5 total **numbers** which are odd = 3 **numbers** which are even = 2, 4, 6 total. Web. Lecture 1 { **Outcomes** , events , and probability MATH-UA.0235 Probability and Statistics To model phenomena which appear to be unpredictable and random, probability theory views them as **outcomes** of an experiment (in the large sense of the word)..

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Web. **The** choices of **outcomes** from one **dice** are six, there are three **dices** so choices for each **dice** will be six but the **outcome** **of** at least one **dice** is made fixed which means at most all **dices** would have a fixed **outcome** which is 5. The **number** **of** **outcomes** which are formed when three **dices** are rolled one after the other but at least one **dice** is fixed at 5.

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My initial reaction is to say that the answer is 6 4, since 4 **dice** can have 6 **outcomes**. **In** my train of thought, the first **dice** can have 6 **outcomes**, same as the second, third and fourth, thus 6 ∗ 6 ∗ 6 ∗ 6 would seem fitting. However, my professor claims that answer is wrong because it overcounts, and says that the answer is actually ( 9 4). **The** **number** **of** **possible** **outcomes** **in** **a** **throw** **of** n **ordinary** **dice** **in** which at least one of the die shows and odd **number** is **a**. 6^n-1 b. 3^n-1 c. 6^n-3^n. So, for example, when we roll one die, we just have six **outcomes** for two **dice**. We have 36 **outcomes**, so we multiply the previous one by six. And so for three **dice**, we now have some multiply 36 by six. Because we have to factor in that. There's six additional faces we have to consider, and that gives us 216 total **outcomes**. And now we can move on.

BITSAT 2020: The **number** **of** **possible** **outcomes** **in** **a** **throw** **of** n **ordinary** **dice** **in** which at least one of the **dice** shows an odd **number** is (**A**) 6n-1 (B) 3n-1.

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Web. Dec 25, 2019 · **The number** **of possible** **outcomes** **in a throw** of n **ordinary** **dice** in which atleast one of the **dice** shows an odd no. - 14310090. **The** **number** **of** **possible** **outcomes** **in** **a** **throw** **of** n **ordinary** **dice** **in** which at least one of the **dice** shows an odd **number** is.

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Solution For The **number** **of** **possible** **outcomes** **in** **a** **throw** **of** \displaystyle{n} **ordinary** **dice** **in** which at least one of the die shows and odd **number** is **a**. \displaystyle{6}^{{n}}-{1} b. \displaystyle{3}^. **The** **possible** **outcomes** are: 1, 2, 3, 4, 5, and 6. The probability of getting any of the **possible** **outcomes** is 1/6. As the possibility of happening any of an event is the same so there are equal chances of getting any likely **number** **in** this case it is either 1/6 or 50/3%. Formula of Probability Since probability is the possibility of an **outcome**. **The number** **of possible** **outcomes** **in a throw** of n **ordinary** **dice** in which at least one of the **dice** shows an odd **number** is (A) 6n-1 (B) 3n-1 (C) 6n-3n (D). **The number** **of possible** **outcomes** **in a throw** of n **ordinary** **dice** in which at least one of the **dice** shows an odd **number** is 🗓 Mar 14, 2022. 3n−1. Web. Solution For **The number** **of possible** **outcomes** **in a throw** of n **ordinary** **dice** in which at least one of the die shows and odd **number** is a. 6^n-1 b. 3^n-1 c. 6^n-3^n d. none of these. Web.

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• Probability of an event E = p(E) = (**number** **of** favorable **outcomes** **of** E)/(number of total **outcomes** **in** **the** sample space) This approach is also called theoretical probability. The example of finding the probability of a sum of seven when two **dice** are tossed is an example of the classical approach.

Nov 26, 2021 · **The number** **of possible** **outcomes** **in a throw** of n **ordinary** **dice** in which at least one of the **dice** shows an odd **number** is A. `6^(n)-1` B. `3^(n)-1` C. `6^(n)-3^(n)` D. `6^(n)-5^(n)` class-12 Share It On FacebookTwitterEmail Please log inor registerto add a comment. 1Answer 0votes answeredSep 6, 2019by EesvarSharma(87.8kpoints). Web. Web.

**The number** **of possible** **outcomes** **in a throw** of n **ordinary** **dice** in which at least one of the **dice** shows an odd **number** is (A) 6n-1 (B) 3n-1 (C) 6n-3n (D).

Web. Web. **The** **number** **of** **possible** **outcomes** **in** **a** **throw** **of** 5 **ordinary** **dice** **in** which at least one of the **dice** shows an odd **number** is A 65−1 B 35 C 65−35 D 55 Solution The correct option is C 65−35 Total **possible** **outcome** =6×6×⋯5 times= 65 The total **number** **of** ways when no odd **outcomes** (only even) **numbers** is 3×3×⋯ to 5 times =35.

Web. Solution Verified by Toppr Correct option is C) Total **number** **of** **possible** **outcomes** **in** **a** **throw** **of** n **ordinary** dies =6×6×6×6×.................n−times=6 n The even **numbers** **in** **the** **throw** **of** **an** **ordinary** **dice** are 2,4,6 The total **number** **of** **possible** **outcomes** when no **dice** show odd **number** =3×3× 3×3×.............n−times=3 n.

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Because the faces are all the same, there is an equal chance of landing on any face. So, using the platonic solids we can have **dice** with 4, 6, 8, 12 or 20 faces! But with some imagination we can actually make fair **dice** with any **number** **of** faces we want. How about these interesting **dice**? Or this 10-sided one.

Nov 26, 2021 · **The number** **of possible** **outcomes** **in a throw** of n **ordinary** **dice** in which at least one of the **dice** shows an odd **number** is A. `6^(n)-1` B. `3^(n)-1` C. `6^(n)-3^(n)` D. `6^(n)-5^(n)` class-12 Share It On FacebookTwitterEmail Please log inor registerto add a comment. 1Answer 0votes answeredSep 6, 2019by EesvarSharma(87.8kpoints).

• Probability of an event E = p(E) = (**number** **of** favorable **outcomes** **of** E)/(number of total **outcomes** **in** **the** sample space) This approach is also called theoretical probability. The example of finding the probability of a sum of seven when two **dice** are tossed is an example of the classical approach. Web. Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

Web. Solution The correct option is C91 Required **number** **of** **possible** **outcomes** = Total **possible** **outcomes**- **possible** **outcomes** **in** winds 5 does not appear = 63−53=91 Suggest Corrections 0 Similar questions Q. If 3dice are rolled, then the **number** **of** **possible** **outcomes** **in** which at least one **dice** shows 5is Q. 4 normal distinguishable **dice** are rolled once.

Web. Web. Because the faces are all the same, there is an equal chance of landing on any face. So, using the platonic solids we can have **dice** with 4, 6, 8, 12 or 20 faces! But with some imagination we can actually make fair **dice** with any **number** **of** faces we want. How about these interesting **dice**? Or this 10-sided one.

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**The** total of points is 21 and the actual corresponding **dice** roll (we have to sum 1 pre-assigned point to each die) would be {2,7,1,5,1,1,4,2,7,1}, with sum 31 but with two outlaw **dice**. Then we arrive at **dice** 9, assign 6 points to it and assign the remaining 15 points to the **dice**. Web. Web.

Web. **The** Question and answers have been prepared according to the CAT exam syllabus. Information about The **number** **of** distinct **outcomes** **possible** with **a** **throw** **of** 3 indistinguishable **dice** isa)216b)36c)56d)72Correct answer is option 'C'.

Solution For The **number** **of** **possible** **outcomes** **in** **a** **throw** **of** \displaystyle{n} **ordinary** **dice** **in** which at least one of the die shows and odd **number** is **a**. \displaystyle{6}^{{n}}-{1} b. \displaystyle{3}^.