Best Answer

There are 16 possible outcomes, each of which is equally likely. So each has a probability of 1/16

HHHH: X = H*T = 4*0 = 0

HHHT: X = H*T = 3*1 = 3

HHTH: X = H*T = 3*1 = 3

HTHH: X = H*T = 3*1 = 3

THHH: X = H*T = 3*1 = 3

HHTT: X = H*T = 2*2 = 4

HTHT: X = H*T = 2*2 = 4

HTTH: X = H*T = 2*2 = 4

THHT: X = H*T = 2*2 = 4

THTH: X = H*T = 2*2 = 4

TTHH: X = H*T = 2*2 = 4

HTTT: X = H*T = 1*3 = 3

THTT: X = H*T = 1*3 = 3

TTHT: X = H*T = 1*3 = 3

TTTH: X = H*T = 1*3 = 3

TTTT: X = H*T = 0*4 = 0

So, the probability distribution function of X is

f(X = 0) = 1/16

f(X = 1) = 4/16 = 1/4

f(X = 2) = 6/16 = 3/8

f(X = 3) = 4/16 = 1/4

f(X = 4 = 1/16

and

f(X = x) = 0 for all other x

Q: How do you find the probability distribution of x if a fair coin is tosses four times and x is the PRODUCT of the number of heads times the number of tails?

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Yes. You are measuring the number of 'successes', x, (in this case the number of heads) out of a number of 'trials', n, (in this case coin tosses) that has an assumed probability, p, (in this case 50% expressed as 0.5) of happening. This phenomenon follows a binomial distribution. Apply the binomial distribution to evaluate whether the the probability of x success from n trials with probability p of occurring is within a pre-determined 'acceptable' limit. Let's say you observe 54 heads in 100 tosses and you wonder if the coin really is fair. From the binomial distribution, the probability of getting *exactly* 54 heads from 100 tosses (assuming that the coin *is* fair & should have 0.5 chance of landing on either side) is 0.0580 or 5.8%. Note that this is not the same probability as 54 heads *in a row*. Most statisticians would agree that 5.8% is too large and conclude that the coin is fair.

The probability of two tails on two tosses of a coin is 0.52, or 0.25.

It's an important principle or probability. The more coin tosses there are, the more chance there is for an expected outcome.

The probability is 1/4

The probability of getting heads on three tosses of a coin is 0.125. Each head has a probability of 0.5. Since the events are sequentially unrelated, simply raise 0.5 to the power of the number of tosses (3) and get 0.125, or 1 in 8.

Related questions

In a large enough number of tosses, it is a certainty (probability = 1). In only the first three tosses, it is (0.5)3 = 0.125

Yes. You are measuring the number of 'successes', x, (in this case the number of heads) out of a number of 'trials', n, (in this case coin tosses) that has an assumed probability, p, (in this case 50% expressed as 0.5) of happening. This phenomenon follows a binomial distribution. Apply the binomial distribution to evaluate whether the the probability of x success from n trials with probability p of occurring is within a pre-determined 'acceptable' limit. Let's say you observe 54 heads in 100 tosses and you wonder if the coin really is fair. From the binomial distribution, the probability of getting *exactly* 54 heads from 100 tosses (assuming that the coin *is* fair & should have 0.5 chance of landing on either side) is 0.0580 or 5.8%. Note that this is not the same probability as 54 heads *in a row*. Most statisticians would agree that 5.8% is too large and conclude that the coin is fair.

The experimental probability of a coin landing on heads is 7/ 12. if the coin landed on tails 30 timefind the number of tosses?

The number of total outcomes on 3 tosses for a coin is 2 3, or 8. Since only 1 outcome is H, H, H, the probability of heads on three consecutive tosses of a coin is 1/8.

2 out of 6

The probability of two tails on two tosses of a coin is 0.52, or 0.25.

It's an important principle or probability. The more coin tosses there are, the more chance there is for an expected outcome.

The probability is 1/4

The probability is 0.0322

The probability is 0, since there will be some 3-tosses in which you get 0, 1 or 3 heads. So not all 3-tosses will give 2 heads.

The probability of getting heads on three tosses of a coin is 0.125. Each head has a probability of 0.5. Since the events are sequentially unrelated, simply raise 0.5 to the power of the number of tosses (3) and get 0.125, or 1 in 8.

The probability is 1 out of 5