Probability is defined as the number of ways an outcome can occur divided by the number of possible outcomes. For the coins, there are 4 outcomes (HH, HT, TH, TT). On the cube, there are 6 possible outcomes. The total number of outcomes is then 4*6 = 24. Since there is only 1 way to obtain HH, look at the cube outcomes. With the HH outcome, the cube would need to fall on a 4. So, there is only 1 way a HH4 can occur. Therefore the probability of getting 2 heads and a four is 1/24 or 0.04167.
There are eight possible outcomes: HHH, HHT, HTT, HTH, TTT, TTH, THH, THT. Of these, 3 contain two tails: HTT, TTH, and THT and the probability of getting two tails is 3/8. If the question were 'getting at least two tails' then TTT would need to be included for a probability of 4/8 or 0.5.
When you toss a coin four times, each toss has 2 possible outcomes: heads or tails. To find the total number of outcomes, you multiply the number of outcomes for each toss together. In this case, it would be 2 x 2 x 2 x 2 = 16 possible outcomes.
With 4 coins you have 24 (16) possibilities. If we wanted a specific coin to land 'tails' while the others landed 'heads' we would have one possibilty out of 16. Since we don't care which of the coins lands 'tails,' we have four chances in 16 or a 25 percent chance.
When tossing a coin, there are 2 distinct possibilities: heads or tails. When tossing 4 coins, there are 2 x 2 x 2 x 2 = 16 possible outcomes. To determine the number of outcomes that will result in 2 heads and 2 tails the formula would be n!/(h!)(n-h)! where n = number of coinsand h = number of heads. In this case the number of different ways you can get 2 heads and 2 tails would be4!/(2! x 2!) = 6 Let's verify the number of different ways you can get two heads and two tails: H H T TH T T HH T H T T T H HT H H TT H T H These are the six different ways to get two heads and two tails with four coins. To find the probability, you divide 6 by the total number of possible outcomes (16) and you would get 6/16 = 3/8. The probability of tossing 4 coins and getting two heads and two tails is 3/8 or 0.375
I believe there would be 11 possible outcomes!
There are eight (8).
Eight of them.
Since each coin would have the outcome with Heads and Tails: Then among the 32 coins, we can have the possible outcomes from no Heads, 1 Head, 2 Heads, ....... , 31 Heads, 32 Heads. Therefore we would have 33 outcomes.
9, you just have to multiply for problems like this
The theoretical probability of HT or TH when two coins are tossed is 1/2 . (All possible outcomes are HH,TT,HT,TH). This means that when we run the experiment repeatedly we expect to get the desired result 1/2 of the time. Since you intend to toss the coins 40 times, 20 are expected.
Given that x is cross and o is head the possible outcomes would be: x-x-x o-x-x x-o-x x-x-o o-o-x o-x-o x-o-o o-o-o Thence there are 8 possible outcomes. Assumptions: 1) Each coin ends up head or tail (never on its side) 2) The coins are distinctinguishable (maybe a penny dime and quarter). If three identical perfectly balanced pennies are tossed at the same time there are 4 outcomes (all heads or all tails 1/8 of the time each and just one or just 2 heads 3/8 of the time each)
Probability is defined as the number of ways an outcome can occur divided by the number of possible outcomes. For the coins, there are 4 outcomes (HH, HT, TH, TT). On the cube, there are 6 possible outcomes. The total number of outcomes is then 4*6 = 24. Since there is only 1 way to obtain HH, look at the cube outcomes. With the HH outcome, the cube would need to fall on a 4. So, there is only 1 way a HH4 can occur. Therefore the probability of getting 2 heads and a four is 1/24 or 0.04167.
Each coin can land in two ways.The die has 6 possible outcomes.So there are 2 x 2 x 6 = 24 possible outcomes for the whole experiment.Note that I am assuming the coins can be told apart - say the first coin and 2nd coin and that H and then T is different that T and then H. If not, then there are only be three outcomes for the coins-- 2 heads, 1 head or no heads and the total number of outcomes would be 3 X 6 = 18.
The single die can land in 6 different ways. For each of those . . . The first coin can land in 2 different ways. For each of those . . . The second coin can land in 2 different ways. Total number of different possible outcomes = (6 x 2 x 2) = 24 IF the coins have different colors so that you can tell them apart. If you can't tell the coins apart, then there are only 12 different outcomes.
2, there could only be two outcomes. Heads or tails As in any experiment you need to define a valid toss. So if you are examining the number of heads and tails you would limit the outcomes to these two states. However, in a real world experiment there may be a condition where the coin landed on a uneven surface and was tilted in some way. The number of tilted ways is infinite. You could also have the coin land on its edge which is unlikely but possible.
There are eight possible outcomes: HHH, HHT, HTT, HTH, TTT, TTH, THH, THT. Of these, 3 contain two tails: HTT, TTH, and THT and the probability of getting two tails is 3/8. If the question were 'getting at least two tails' then TTT would need to be included for a probability of 4/8 or 0.5.