A 12-sided die is rolled. The set of equally likely outcomes is {1,2,3,4,5,6,7,,8,9,10,11,12}. Find the probability of rolling a number less than 6. We want a {1, 2, 3, 4, 5} P(X < 6) =[B] 5/12[/B]

A 12-sided die is rolled. The set of equally likely outcomes is {1,2,3,4,5,6,7,8,9,10,11,12}. Find the probability of rolling a number less than 6. We have 12 outcomes. Less than 6 means 1, 2, 3, 4, 5. Our probability P(x < 6) is: P(x < 6) = [B]5/12[/B]

A coin is tossed 3 times. a. Draw a tree diagram and list the sample space that shows all the possible outcomes [URL='https://www.mathcelebrity.com/cointoss.php?hts=+HTHTHH&hct=+2&tct=+1&fct=+5>=no+more+than&nmnl=+2&htpick=heads&tossct=3&montect=3&calc=5&pl=Calculate+Probability']type in "toss a coin 3 times" and pick the probability option[/URL].

A fair coin is tossed 4 times. a) How many outcomes are there in the sample space? b) What is the probability that the third toss is heads, given that the first toss is heads? c) Let A be the event that the first toss is heads, and B be the event that the third toss is heads. Are A and B independent? Why or why not? a) 2^4 = [B]16[/B] on our [URL='http://www.mathcelebrity.comcointoss.php?hts=+HTHTHH&hct=+2&tct=+1&fct=+5>=no+more+than&nmnl=+2&htpick=heads&tossct=+4&calc=5&montect=+500&pl=Calculate+Probability']coin toss calculator[/URL] b) On the link above, 4 of those outcomes have H and H in toss 1 and 3. So it's [B]1/4 or 0.25[/B] c) [B]Yes, each toss is independent of each other.[/B]

[SIZE=6][B]A quiz has 5 questions with 4 answer choices each find the number of possible outcomes[/B] [B][/B] [B]We have 4 * 4 * 4 * 4 * 4 = 1024 outcomes[/B][/SIZE]

A number cube is rolled and a coin is tossed. The number cube and the coin are fair. What is the probability that the number rolled is greater than 3 and the coin toss is heads? Write your answer as a fraction in simplest form Let's review the vitals of this question: [LIST] [*]The probability of heads on a fair coin is 1/2. [*]On a fair die, greater than 3 means either 4, 5, or 6. Any die roll face is a 1/6 probability. [*]So we have a combination of outcomes below: [/LIST] Outcomes [LIST=1] [*]Heads and 4 [*]Heads and 5 [*]Heads and 6 [/LIST] For each of the outcomes, we assign a probability. Since the coin flip and die roll are independent, we multiply the probabilities: [LIST=1] [*]P(Heads and 4) = 1/2 * 1/6 = 1/12 [*]P(Heads and 5) = 1/2 * 1/6 = 1/12 [*]P(Heads and 6) = 1/2 * 1/6 = 1/12 [/LIST] Since we want any of those events, we add all three probabilities 1/12 + 1/12 + 1/12 = 3/12 This fraction is not simplified. S[URL='https://www.mathcelebrity.com/fraction.php?frac1=3%2F12&frac2=3%2F8&pl=Simplify']o we type this fraction into our search engine, and choose Simplify[/URL]. We get a probability of [B]1/4[/B]. By the way, if you need a decimal answer or percentage answer instead of a fraction, we type in the following phrase into our search engine: [URL='https://www.mathcelebrity.com/perc.php?num=1&den=4&pcheck=1&num1=+16&pct1=+80&pct2=+35&den1=+90&pct=+82&decimal=+65.236&astart=+12&aend=+20&wp1=20&wp2=30&pl=Calculate']1/4 to decimal[/URL] Alternative Answers: [LIST] [*]For a decimal, we get [B]0.25[/B] [*]For a percentage, we get [B]25%[/B] [/LIST]

A pair of standard dice is rolled, how many possible outcomes are there? We want the number of outcomes in the sample space. The first die has 6 possibilities 1-6. The second die has 6 possibilities 1-6. Our sample space count is 6 x 6 = [B]36 different outcomes [/B] [LIST=1] [*](1, 1) [*](1, 2) [*](1, 3) [*](1, 4) [*](1, 5) [*](1, 6) [*](2, 1) [*](2, 2) [*](2, 3) [*](2, 4) [*](2, 5) [*](2, 6) [*](3, 1) [*](3, 2) [*](3, 3) [*](3, 4) [*](3, 5) [*](3, 6) [*](4, 1) [*](4, 2) [*](4, 3) [*](4, 4) [*](4, 5) [*](4, 6) [*](5, 1) [*](5, 2) [*](5, 3) [*](5, 4) [*](5, 5) [*](5, 6) [*](6, 1) [*](6, 2) [*](6, 3) [*](6, 4) [*](6, 5) [*](6, 6) [/LIST]

A spinner has 3 equal sections labelled A, B, C. A bag contains 3 marbles: 1 grey, 1 black, and 1 white. The pointer is spun and a marble is picked at random. a) Use a tree diagram to list the possible outcomes. [LIST=1] [*][B]A, Grey[/B] [*][B]A, Black[/B] [*][B]A, White[/B] [*][B]B, Grey[/B] [*][B]B, Black[/B] [*][B]B, White[/B] [*][B]C, Grey[/B] [*][B]C, Black[/B] [*][B]C, White[/B] [/LIST] b) What is the probability of: i) spinning A? P(A) = Number of A sections on spinner / Total Sections P(A) = [B]1/3[/B] --------------------------------- ii) picking a grey marble? P(A) = Number of grey marbles / Total Marbles P(A) = [B]1/3[/B] --------------------------------- iii) spinning A and picking a white marble? Since they're independent events, we multiply to get: P(A AND White) = P(A) * P(White) P(A) was found in i) as 1/3 Find P(White): P(White) = Number of white marbles / Total Marbles P(White) = 1/3 [B][/B] Therefore, we have: P(A AND White) = 1/3 * 1/3 P(A AND White) = [B]1/9[/B] --------------------------------- iv) spinning C and picking a pink marble? Since they're independent events, we multiply to get: P(C AND Pink) = P(C) * P(Pink) Find P(C): P(C) = Number of C sections on spinner / Total Sections P(C) = 1/3 [B][/B] Find P(Pink): P(Pink) = Number of pink marbles / Total Marbles P(Pink) = 0/3 [B][/B] Therefore, we have: P(C AND Pink) = 1/3 * 0 P(C AND Pink) = [B]0[/B]

can 0.2 be the probability of an outcome in a sample space? Yes. Any probability p is a valid sample space outcome if: [B]0 <= p <= 1[/B]

Can a coefficient of determination be negative? Why or why not? [B]Yes, reasons below[/B] [LIST] [*] predictions that are being compared to the corresponding outcomes have not been derived from a model-fitting procedure using those data [*] where linear regression is conducted without including an intercept [*] Yes, negative values of R2 may occur when fitting non-linear functions to data [/LIST]

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flip 7 coins How many total outcomes are there A flip of a coin has 2 outcomes, heads or tails. Since each outcome is independent of the other outcomes, we multiply each flip by 2 outcomes: Total outcomes = 2 * 2 * 2 * 2 * 2 * 2 * 2 Total outcomes = 2^7 Total outcomes = [B]128[/B]

How many possible outcomes of rolling 4 dice Each die has 6 faces, so we have: 6 * 6 * 6 * 6 = [B]1,296 possible outcomes[/B]

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sample space for flipping a coin 3 times Each flip gives us 2 possible outcomes, heads or tails. So we have: 2 * 2 * 2 = 8 possible outcomes [LIST=1] [*]HHH [*]HHT [*]HTH [*]HTT [*]THH [*]THT [*]TTH [*]TTT [/LIST]

List out the sums greater than 8: (4, 5) (4, 6) (5, 5) (5, 6) (6, 6) (5, 4) (6, 4) (6, 5) Since there are 6 * 6 = 36 total outcomes, we have the probability of the sum greater than 8 as: 8/36 = 2/9

Ted tossed a number cube and rolled a die. How many possible outcomes are there? A number cube has 6 possible outcomes A die has 6 possible outcomes. We have 6 * 6 = [B]36 possible outcomes[/B].

the sample space for a coin being tossed twice Since each toss results in 2 outcomes, we have 2^2 = 4 possible events in the sample space: [LIST=1] [*]H,H [*]H,T [*]T,H [*]T,T [/LIST]

three coins are tossed.how many different ways can they fall? [URL='https://www.mathcelebrity.com/cointoss.php?hts=+HTHTHH&hct=+2&tct=+1&fct=+5>=no+more+than&nmnl=+2&htpick=heads&tossct=3&montect=3&calc=5&pl=Calculate+Probability']8 outcomes using our coin toss calculator[/URL]

Two coins are flipped 2 times. Calculate the total outcomes of these coins. 2 coins * 2 outcomes per coin = 4 possible outcomes [LIST=1] [*][B]H,H[/B] [*][B]H,T[/B] [*][B]T,H[/B] [*][B]T,T[/B] [/LIST]

What is the sample space for a 10 sided die? Sample space means the set of all possible outcomes. For a 10-sided die, we have: [B]{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}[/B]

Write a sample space for rolling a dice twice Each die roll has 6 possible outcomes. So 2 die-rolls has 6^2 = 36 possible outcomes: [LIST=1] [*]1,1 [*]1,2 [*]1,3 [*]1,4 [*]1,5 [*]1,6 [*]2,1 [*]2,2 [*]2,3 [*]2,4 [*]2,5 [*]2,6 [*]3,1 [*]3,2 [*]3,3 [*]3,4 [*]3,5 [*]3,6 [*]4,1 [*]4,2 [*]4,3 [*]4,4 [*]4,5 [*]4,6 [*]5,1 [*]5,2 [*]5,3 [*]5,4 [*]5,5 [*]5,6 [*]6,1 [*]6,2 [*]6,3 [*]6,4 [*]6,5 [*]6,6 [/LIST]

You roll a red die and a green die. What is the size of the sample space of all possible outcomes of rolling these two dice, given that the red die shows an even number and the green die shows an odd number greater than 1? [LIST] [*]Red Die Sample Space {2, 4, 6} [*]Green Die Sample Space {3, 5} [*]Total Sample Space {(2, 3), (2, 5), (4, 3), (4, 5), (6, 3), (6, 5)} [*]The sie of this is 6 elements. [/LIST]