A coin is tossed five times. Find the probability of getting exactly three heads.?

A coin is tossed five times. Find the probability of getting exactly three heads.?

You have to get three heads and two tails, and there is more than one way of doing this. The formula isp(3) = 5C3 × 0.5^3 × 0.5^2where 5C3 = 5!/3!2! = 10 is the number of ways of selecting 3 things out of 5.This gives the probability of exactly three heads as 0.3125..

The probability of getting a head in each toss is 1/2. Similarly the probability of getting a tail is also 1/2. The probability distribution is binomial.(H + T)^5 and the coefficient of the term H^3 will be the probability.

The probability is always 50 - 50. Chance confounds the law of averages because each flip is individual, each rotation of the coin will be unlike the previous or the next flip. It is just as likely to produce five "heads" as five "tails". Unless, of course, you know what you're doing. But that's a route that eliminates chance.

Tougher than it looks. Here we go:Flip a coin. You get H (heads) or T (tails). So two possible outcomes in one flip. If you flip it 5 times, you have 2^5=32 possible outcomes.How many of these 32 outcomes contain exactly 3 heads?When we have three heads, we must also have exactly three tails, so your goal is to determine how many combinations of this there are. Consider one option:HHHTTFirst we need to flip three heads in a row. The odds of this happening are (1/2)^3=1/8Then we need to flip two tails in a row. Odds: (1/2)^2=1/4. So the probability of getting HHHTT is (1/8)(1/4)=1/32 (good general idea to remember in probability problems: x=and, +=or).Now consider another option: HHTHT. We flip 2 heads (odds: (1/2)^2=1/4), 1 tails (odds: 1/2), 1 heads (1/2), and another tails (1/2). So the odds of getting this combination are (1/4)(1/2)(1/2)(1/2)=1/32.The point is, the probability of getting each combination of 3 H's and 2 T's is always the same, 1/32. So, how many possible 5 flip combinations have 3 heads? Well, that would be solved as a combination, 5C3=5!/(3!2!)=10. So the answer is 10(1/32)=10/32=5/16

Flip a coin. You get H (heads) or T (tails). So two possible outcomes in one flip. If you flip it 5 times, you have 2^5=32 possible outcomes.How many of these 32 outcomes contain exactly 3 heads?When we have three heads, we must also have exactly three tails, so your goal is to determine how many combinations of this there are. Consider one option:HHHTTFirst we need to flip three heads in a row. The odds of this happening are (1/2)^3=1/8Then we need to flip two tails in a row. Odds: (1/2)^2=1/4. So the probability of getting HHHTT is (1/8)(1/4)=1/32 (good general idea to remember in probability problems: x=and, +=or).Now consider another option: HHTHT. We flip 2 heads (odds: (1/2)^2=1/4), 1 tails (odds: 1/2), 1 heads (1/2), and another tails (1/2). So the odds of getting this combination are (1/4)(1/2)(1/2)(1/2)=1/32.The point is, the probability of getting each combination of 3 H's and 2 T's is always the same, 1/32. So, how many possible 5 flip combinations have 3 heads? Well, that would be solved as a combination, 5C3=5!/(3!2!)=10. So the answer is 10(1/32)=10/32=5/16 this is wat my teacher said.

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