### 4 Answers

- PuzzlingLv 72 months ago
There are 8 possible outcomes.

HHH, HHT, HTH, HTT, THH, THT, TTH, TTT

Just count the number of desired outcomes and divide by 8.

PROBLEM 1:

3 heads --> HHH

P(3 heads) = 1/8

PROBLEM 2:

2 heads and 1 tail --> HHT, HTH, THH

P(2 heads and 1 tail) = 3/8

PROBLEM 3:

At least one tail --> HHT, HTH, HTT, THH, THT, TTH, TTT

P(at least one tail) = 7/8

P.S. A faster way to figure this might be to think of the opposite case. If you don't toss at least one tail, it means you tossed all 3 heads. We know the probability of 3 heads is 1/8, so the probability of no heads (aka at least one tail) is:

1 - 1/8

= 7/8

PROBLEM 4:

At least two heads --> HHH, HHT, HTH, THH

P(at least two heads) = 4/8 = 1/2

P.S. The two cases we are considering are 3 heads (0 tails) and 2 heads (1 tail). But by symmetry, we could be counting the same thing for tails with 3 tails (0 heads) and 2 tails (1 head). These two probabilities will be exactly the same and that covers all the possible outcomes. As a result, each probability will be the same = 1/2.

P.P.S. You could also add the results for part 1 and part 2.

P(at least 2 heads) = P(3 heads) + P(2 heads)

= 1/8 + 3/8

= 4/8

= 1/2

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- Engr. RonaldLv 72 months ago
1. 3 heads

P( 3 heads) = 1/8 answer//

2. 2 heads and tail

P(2H and T) = 3/8 answer//

3. At least 1 tail

P(at least 1 tail) = 7/8 answer//

3. At least 2 heads

P(at least 2 heads) = 4/8 or 1/2 answer..//

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- 冷眼旁觀Lv 72 months ago
H: Head

T: Tail

P(H) = 1/2

P(T) = 1/2

====

1.

P(3H)

= P(H)³

= (1/2)³

= 1/8

= 0.125

====

2.

Method 1:

P(2H 1T)

= P(HHT) + P(HTH) + P(THH)

= (1/2)³ + (1/2)³ + (1/2)³

= 3/8

= 0.375

Method 2:

P(2H 1T)

= ₃C₂ × (1/2)² × (1/2)

= 3 × (1/4) × (1/2)

= 3/8

= 0.375

====

3.

P(at least 1T)

= 1 - P(3T)

= 1 - (1/2)³

= 1 - (1/8)

= 7/8

= 0.875

====

4.

P(at least 2H)

= P(2H 1T) + P(3H)

= 0.375 + 0.125

= 0.5

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