th
vq
Enterprise

The number of possible outcomes in a throw of an ordinary dice

ni

A hand ringing a receptionist bell held by a robot hand

Math Probability Q&A Library Exercise 1 An ordinary dice is rolled. The number 5 shows. 1.1 What is the experiment in this case? 1.2 What is the outcome of the experiment in this case? 1.3 List the possible outcomes of this experiment.

pm
ip

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.

Web.

Web.

av

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`.

Web.

Web.

Web.

Web.

jo

(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)..

Web. Web.

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.

Web.

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.

ai

Web. Web.

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.

fy

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.

Web.

Web.

• 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.

sz

ir
al
cq

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.

xn

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.

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.

Web.

Web.

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}^.

je
ti
Policy

qt

kx

Mar 25, 2016 · Consider you throw 2 dices. How many outcomes do you have? If they are distinguishable (die1, die2) then clearly 36 options (any number of die 1 and any number of die2). If you cannot distinguish them, then you have less options, as ( ( 1, 4) = ( 4, 1)) etc. – Peter Franek Mar 25, 2016 at 9:09 n+5 choose 5 is the indistinguishable option..

gn

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.

mj wv
xm
az

Web. Web. . Web. 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'..

iv

jc

1. Fill the carafe with water according to the number of cups of coffee you need to make. 2. Pour the water from the carafe into the reservoir of the coffee maker, and place the carafe back into position. 3. Place a coffee filter into the filter basket. The amount of coffee you'll need to add depends on how strong or weak your clients like it .... Jan 06, 2020 · The number of possible outcomes in a throw of n n ordinary dice in which at least one of the die shows and odd number is a. 6n − 1 6 n - 1 b. 3n − 1 3 n - 1 c. 6n − 3n 6 n - 3 n d. none of these. A. 6n − 1 6 n - 1. B. 3n − 1 3 n - 1. C. 6n − 3n 6 n - 3 n. D. none of these..

7 Intuitively we would expect the sum of a single die to be the average of the possible outcomes, ie: S= (1+2+3+4+5+6)/6 = 3.5 And so we would predict the sum of a two die to be twice that of one die, ie we would predict the expected value to be 7 If we consider the possible outcomes from the throw of two dice: And so if we define X as a random variable denoting the sum of the two dices, then.

px xk
lu
da

In a simultaneous throw of two dice, we have n(S)=(6⋅6)=36. What is the probability of getting a doublet when a die is thrown twice? We know that the favourable outcomes are (2, 2), (4, 4), and (6, 6). So, the number of favourable outcomes is 3. We know that the number of possible outcomes is 36..

hi ke
Fintech

xs

op

el

rg

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'. Verified by Toppr Correct option is C) Since dice are identical so number of arrangements are possible like 2,3,4,5,6,1 and 1,3,2,4,5,6 will be considered as same outcome. So number of possible outcome is simply the number of terms in the expansion of (x+x 2+x 3+x 4+x 5+x 6) n. That is n+6−1C 6−1.

Write the total number of possible outcomes in a throw of 3 dice in which at least one of the dice shows an even number.📲PW App Link - https://bit.ly/YTAI_P.... Web.

wm st
vl
kj
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. Finding Number of possible choices A coin tossed has two possible outcomes, showing up either a head or a tail. ⇒ The number of possible choices in tossing a coin = 2 Total Event (E) The event of tossing the first of the coins 1st sub-event (SE1) The event of tossing the first of the coins This event can be accomplished in 2 ways ⇒ n SE1 = 2.
cn

The total number of ways is 6 × 6 × ... to n times = 6 n. The total number of ways to show only even numbers is 3 × 3 × . to n times = 3 n . Therefore, the required number of ways is 6 n - 3 n ..

qb

The most common roll is therefore seen to be a 7, with probability , and the least common rolls are 2 and 12, both with probability 1/36. For six-sided dice, and For three six-sided dice, the most common rolls are 10 and 11, both with probability 1/8; and the least common rolls are 3 and 18, both with probability 1/216.

Web. Web. 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.

ds sf
md
vb

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'.

Enterprise

cq

iy

nh

hk

yg

Web. 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).

jp fx
qr
dc

The number of possible outcomes in a throw of \\( \\mathrm{n} \\) ordinary dice in which at least one of the dice shows an odd number is(a) \\( 6^{n}-1 \\)(b) \\( ....

dc
ik
zh
oi
ee
jt
kg
ro