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Q: Find the odds for getting two tails when two fair coins are flipped?

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The probability of flipping one coin and getting tails is 1/2. In order to find the probability of multiple events occurring, you find the product of all the events. For 3 coins the probability of getting tails 3 times is 1/8 because .5 x .5 x .5 = .125 or 1/8.

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

It is 1/2.

A random sample of size 36 is taken from a normal population with a known variance If the mean of the sample is 42.6. Find the left confidence limit for the population mean.

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The probability of flipping one coin and getting tails is 1/2. In order to find the probability of multiple events occurring, you find the product of all the events. For 3 coins the probability of getting tails 3 times is 1/8 because .5 x .5 x .5 = .125 or 1/8.

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

It is 1/2.

You can find a 'theoretical probability' or a 'mathematical probability' witha pencil and paper. But the only way to find an experimental probabilityis to do the experiment.(Also, before you do the experiment, you really need to define the 'successfuloutcome' a little more clearly. Like, what does "head and one tails" mean, howmany coins are being flipped for each trial, and how many trials will there be ? )

A random sample of size 36 is taken from a normal population with a known variance If the mean of the sample is 42.6. Find the left confidence limit for the population mean.

Read it and find out.

Chances of getting one tail is 50:50, which equals to .5. In order to find out about four coins you have to multiply .5 x .5 x .5 x .5 which is 6.25 % - a very low chance of success. Same is valid if you were flipping the exact same coin, since each event (flip) is independent and unrelated to the previous.

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

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First work out the probability of the first two coins being heads and then the last being tails. This is 1/2 x 1/2 x 1/2 which is 1/8 The next step is to find out how many different orders the coins can come in. In this case there are 3 possible orders (HHT, HTH, and THH). Multiply this by the above probability and you get 3/8. Therefore the probability of getting two heads and one tail is 3/8

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