Probability

# What is the possible outcomes when two coins are tossed?

Tails ( T )

HH , TT , HT , TH

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## Related Questions

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

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.

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.

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

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.

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

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

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.

One die, two dice and no such word as dices! To chart the possible outcomes consider the set of ordered pairs (X, Y) where X is the outcome for the first die and Y is the outcome for the second die.

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.

It depends on the definition of an outcome. If you care about the order of the tosses, &lt;br /&gt; you get 2 possible outcomes per toss. Three tosses give you 2*2*2=8 possible outcomes. If you only care about the final number of heads and tails, there are 4 possible outcomes (3 heads, 2 heads and a tail, a head and two tails, or 3 tails).

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.

Two coins tossed sample space is (H=Heads, T = Tails) as follows. HH, HT, TH, TT is the sample space.

Possible outcomes are HH, HT, TH, TT; therefore the probability of HH is 1/4 or 0.25.

There are many more than two possible outcomes in a civil case. Your question is too broad.

At least two heads with two coins? You can't get more.There are 4 different outcomes:tail-tail, head-tail, tail-head and head-head.You can use one out of four - which gives us the probability 1/4 = 0.25 = 25%

Counting Principle is used to find the number of possible outcomes. It states that if an event has m possible outcomes and another independent event has n possible outcomes, then there are mn possible outcomes for the two events together.

It is the event that one of the two coins lands showing tails and the other shows heads.

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