Math and Arithmetic

Statistics

Probability

Top Answer

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 .

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0When 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.

Each toss has 2 outcomes; so the number of outcomes for 3 tosses is 2*2*2 = 8

The probability of 2 coins both landing on heads or both landing on tails is 1/2 because there are 4 possible outcomes. Head, head. Head, tails. Tails, tails. Tails, heads. Tails, heads is different from heads, tails for reasons I am unsure of.

If they are fair coins, it is 1/16.

the outcomes are 50:50.

A probability is fair if there is no bias in any of the possible outcomes. Said another way, all of the possible outcomes in a fair distribution have an equal probability.

There are 36 possible outcomes of which 6 show the same number on each, so the odds are 5 - 1 against.

Assuming order is irrelevant, 2^5, or 2*2*2*2*2 or 32 possible combos.

The probability is 3/8 = 0.375

With 5 coin tosses there are 32 possible outcomes. 10 of these have exactly 2 heads, and 26 of these have 2 or more heads.For exactly two coins are heads: 10/32 = 31.25%For two or more heads: 26/32 = 81.25%

Assuming the coins are fair, two-sided coins, and landing on their sides is not an option, there are four possible outcomes if you consider coin a having a head and coin b having a tail being a different instance from coin a being a tail and coin be having a head. Here they are; Coin A | Coin B Heads | Tails Heads | Heads Tails....| Heads Tails....| Tails

There is 24 or 16 outcomes. There is 4 ways to get heads once (HTTT & THTT & TTHT & TTTH). So, the probability of getting heads only once if a fair coin is tossed 4 times is 4/16 or 1/4.

Heads or tails; each have a probability of 0.5 (assuming a fair coin).

6 outcomes for each die and so 6x6x6 outcomes for all three

Number of different possible outcomes on a fair die = 6 . Number of outcomes that are less than 8 = 6 . The probability of success is 6/6 = 1.00 = 100% .

In three flips of a fair coin, there are a total of 8 possible outcomes: T, T, T; T, T, H; T, H, T; T, H, H; H, H, H; H, H, T; H, T, H; H, T, T Of the possible outcomes, four of them (half) contain at least two heads, as can be seen by inspection. Note: In flipping a coin, there are two possible outcomes at each flipping event. The number of possible outcomes expands as a function of the number of times the coin is flipped. One flip, two possible outcomes. Two flips, four possible outcomes. Three flips, eight possible outcomes. Four flips, sixteen possible outcomes. It appears that the number of possible outcomes is a power of the number of possible outcomes, which is two. 21 = 2, 22 = 4, 23 = 8, 24 = 16, .... Looks like a pattern developing there. Welcome to this variant of permutations.

Depends if the dice is bias then you an not have an answer if it is fair then you times 6 by 3 which = 18 so if you want three different outcomes then 3/18 which is simplified to 1/6 of a chance!

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.

each toss is independent of the previous one(s), so both outcomes are equiprobable.

First, you find the probability of picking 1 in 5 disregarding order (combination). 5 C 1 = 5/1 = 5 Then you divide that by the possible number of outcomes which is 25. 5/ 2^5= .0195

H H T T H T T H 1 in 2 chances

Sample space: {hhh,hht,hth,htt,ttt,tth,tht,thh} 8 possible outcomes

When a fair die is rolled, there are 6 possible outcomes {1,2,3,4,5,6}. The sample space consists of 6 points, so its size is 6.

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