This is a good bet to take. Your expected payout is 0.5 each round of the 2 tosses. The possible outcomes from 2 tosses: HT, HH, TT, TH. The probability that heads comes up is 3 in 4 (.75). The probability that heads does not come up is 1 in 4 (.25). Your expected payout is: (2 * .75) + (-4 * .25) = 1.5 - 1 = 0.5
It is 1/4.
2:3...
There is a .25 chance that no heads will come up. (1/2*1/2=.25) Therefore, there is a .75 chance that one or more heads will come up. Value of game = .75*$2-.25*$4=$1.5-$1=$.50 Over time, you should come out ahead.
Every time a coin is tossed there is a 50 / 50 chances of it coming up heads. There is no rule that says tossing it 100 or 6 times will change this.
The outcome space is TT, TH, HT, HH: 4 outcomes in all. {One head} = {TH, HT} with prob = 2/4 = 1/2 {No heads} = {TT} with prob 1/4 So E = 2*Pr(one H) - 4*Pr(no H) = 2*(1/2)-4*(1/4) = 1 - 1 = 0
It is 1/4.
The answer depends on how many times the coin is tossed. The probability is zero if the coin is tossed only once! Making some assumptions and rewording your question as "If I toss a fair coin twice, what is the probability it comes up heads both times" then the probability of it being heads on any given toss is 0.5, and the probability of it being heads on both tosses is 0.5 x 0.5 = 0.25. If you toss it three times and want to know what the probability of it being heads exactly twice is, then the calculation is more complicated, but it comes out to 0.375.
2:3...
1/2 x 1/2 = 1/4
There is a .25 chance that no heads will come up. (1/2*1/2=.25) Therefore, there is a .75 chance that one or more heads will come up. Value of game = .75*$2-.25*$4=$1.5-$1=$.50 Over time, you should come out ahead.
Every time a coin is tossed there is a 50 / 50 chances of it coming up heads. There is no rule that says tossing it 100 or 6 times will change this.
The probability is 0.25.Look at it this way--if you toss a coin twice, there are four equally-probable outcomes:tails, tailstails, headsheads, tailsheads, headsSo the probability of heads twice in a row is one in four, or 25%.the chance of tossing heads is 1/2 (50%) The chance of tossing the next heads is 1/2 (50%) 1/2 x 1/2 = 1/4 (25%)
The outcome space is TT, TH, HT, HH: 4 outcomes in all. {One head} = {TH, HT} with prob = 2/4 = 1/2 {No heads} = {TT} with prob 1/4 So E = 2*Pr(one H) - 4*Pr(no H) = 2*(1/2)-4*(1/4) = 1 - 1 = 0
7878
One in four. 1:4. The probability of getting heads when a fair coin is tossed is: P(H) = 1/2. The probability of getting heads on a second toss is: P(H) = 1/2, this result is independent of the result of the first toss. The probability of having both events happen (heads on the first and heads on the second toss) is: P(H1UH2) = (1/2)∙(1/2) = 1/4 = 0.25 = 0.25%
75%
1/4