Math and Arithmetic

Statistics

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Probability

Top Answer

I believe there would be 11 possible outcomes!

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08 outcomes are possible in this situtation. You just have to multiply 4 by 2 to get the answer.

The possible outcomes areHH, HT, TH and TT.

If you know which coin is which, there are 16possible outcomes.If you're only counting the number of Heads and Tails, there are 5 .

Heads ( H ) Tails ( T ) HH , TT , HT , TH

There are eight possible outcomes: HHH, HHT, HTT, HTH, THT, TTT, TTH, THH.

The total number of outcomes is 2^5 = 32.

9, you just have to multiply for problems like this

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.

There are 48 possible outcomes and I have no intention of listing them all. They are all of the form CCCD where C = H or T, and D takes the numeric values from 1 to 6.

There are 25 or 32 possible outcomes can you get by tossing 5 coins.

Each coin has two possible outcomes, either Heads or Tails. Then the number of outcomes when all 4 coins are tossed is, 2 x 2 x 2 x 2 = 16.

When two fair coins are tossed, you have the following possible outcomes: HH, HT, TH, TT. So, at most implies that you get either i) zero heads or ii) one head. From the possible outcomes we see that 3 times we satisify the outcome. Thus, probability of at most one head is 3/4.

There are 23 = 8 possible outcomes.

There are technically 8 possible outcomes if you are talking about the side of the coin it lands on. Each coin has 2 possible outcomes (landing on heads and landing on tails). To figure out the number of outcomes for all the coins you multiply the outcomes for all of the coins together: 2 X 2 X 2= 8.

Let's call one coin A and the other B. omes The possible outcomes for the coins are; A heads and B tails, A tails and B heads, A and B heads, A and B tails. That's four outcomes. The possible outcomes for a single die (as in dice) are six since a die has six faces, So four times six is twenty four possible outcomes.

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.

If you toss eight coins, there are 256 (28) different outcomes.

Use Pascal's Triangle Answer is 14641 different outcomes. - - - - 1 - - - 1 - 1 - - 1 - 2 - 1 - 1 - 3 - 3 - 1 1 - 4 - 6 - 4 - 1

Assuming the variable of interest is the face on top: H (= heads) or T (= tails), then they are the four possible outcomes: HH, HT, TH and TT.

The probability of it is 37.5%. Since there are 16 possible outcomes. calculated by 2 to the power of 4. only 6 gives you the prefered results then it's a simple 6:16 ratio

This question can be rather easily answered, as soon as outcomes 'a' and 'b' are defined.

Number of possible outcomes of one coin = 2Number of possible outcomes of six coins = 2 x 2 x 2 x 2 x 2 x 2 = 64Number of possible outcomes with six heads = 1Probability of six heads = 1/64Probability of not six heads = at least one tails = 63/64 = 98.4375%

There are two outcomes for each coin and three coins; 2 x 2 x 2 = 23 = 8 outcomes.

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.

For each of the coins, in order, you have two possible outcomes so that there are 2*2*2*2 = 16 outcomes in all.

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