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Three coins are tossed, and the number of heads observed is recorded. (Give your answers as fractions.)
(a) Find the probability for 0 heads.

(b) Find the probability for 1 head.

(c) Find the probability for 2 heads.

(d) Find the probability for 3 heads.
3/3 or 1 because of three coins and tossed three times...

On each toss, there is a 1/2 probability of a head.
(a) and (d) are the same, because p(H) = p(T) = 1/2 so 3 heads is the same as 3 tails (0 heads)

The events are independent, so P(HHH) = P(H)*P(H)*P(H) = 1/2 * 1/2 * 1/2 = 1/8

(a) and (d) are both 1/8

(b) and (c) are the same, since P(1 head) = P(1 tail) = P(2 heads)

P(1 head) = P(HTT)+P(THT)+P(TTH) = 1/8 + 1/8 + 1/8 = 3/8
or,
P(1 head) = (1-P(HHH)-P(TTT))/2 = (1 - 2/8)/2 = 3/8

Each specific ordering of throws has a 1/8 chance of coming up, since there are 8 possible outcomes.

If you lst the possible outcomes, you will see that 1 has all heads, 1 has all tails, and 3 have 2 of one and 1 of the other.

You should never have guessed 0/3 on (a) since there is obviously a chance for a head on every throw.

I see what you are saying now, I even tossed a coin trying to figure this out. Thank you, some of these really get to me but I get some of them right off.

Tossing a coin was a good idea, but you can't control what happens. Better would have been to take three coins and line them up, then ask

what are all the possible outcomes of 3 coins in a row? and start flipping them around.

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