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Draw a tree diagram for three fair coins tossed together once. a) Derive the sample space. b) Find the probability of getting: i) at least one head ii) at most one tail iii) exactly three heads.

Mathematics

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Answers

3 parts
b(i)

78\frac{7}{8}

b(ii)

12\frac{1}{2}

b(iii)

18\frac{1}{8}

Step-by-step solution

6 steps

Here's the breakdown for tossing three fair coins:

Part a

Draw a tree diagram for three fair coins tossed together once. Derive the sample space.

Tree Diagram Description:
Start with a single point.

  1. From the starting point, draw two branches for the first coin: one labeled 'H' (Heads) and one labeled 'T' (Tails).
  2. From the end of each of these branches, draw two more branches for the second coin: 'H' and 'T'. This gives paths like HH, HT, TH, TT.
  3. From the end of each of these four branches, draw two final branches for the third coin: 'H' and 'T'. This will lead to eight final outcomes.

Sample Space:
The sample space (S) lists all possible outcomes when three fair coins are tossed.

S={HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}

The total number of possible outcomes is n(S)=8n(S) = 8.

Part b(i)

78\frac{7}{8}

at least one head

Find the probability of getting

  1. Identify outcomes with at least one head.

    This means outcomes with one, two, or three heads.
    The outcomes are: $HHH, HHT, HTH, HTT, THH, THT, TTH$.
    The number of favorable outcomes is 7.

  2. Calculate the probability.

    P(at least one head)=Number of outcomes with at least one headTotal number of outcomes=78P(\text{at least one head}) = \frac{\text{Number of outcomes with at least one head}}{\text{Total number of outcomes}} = \frac{7}{8}

    The probability of getting at least one head is 78\boxed{\frac{7}{8}}.

Part b(ii)

12\frac{1}{2}

at most one tail

  1. Identify outcomes with at most one tail.

    This means outcomes with zero tails or one tail.

    • Outcomes with zero tails: HHHHHH (1 outcome)
    • Outcomes with one tail: HHT,HTH,THHHHT, HTH, THH (3 outcomes) The total number of favorable outcomes is 1+3=41 + 3 = 4.

  2. Calculate the probability.

    P(at most one tail)=Number of outcomes with at most one tailTotal number of outcomes=48=12P(\text{at most one tail}) = \frac{\text{Number of outcomes with at most one tail}}{\text{Total number of outcomes}} = \frac{4}{8} = \frac{1}{2}

    The probability of getting at most one tail is 12\boxed{\frac{1}{2}}.

Part b(iii)

18\frac{1}{8}

exactly three heads

  1. Identify outcomes with exactly three heads.

    The only outcome with exactly three heads is HHHHHH.
    The number of favorable outcomes is 1.

  2. Calculate the probability.

    P(exactly three heads)=Number of outcomes with exactly three headsTotal number of outcomes=18P(\text{exactly three heads}) = \frac{\text{Number of outcomes with exactly three heads}}{\text{Total number of outcomes}} = \frac{1}{8}

    The probability of getting exactly three heads is 18\boxed{\frac{1}{8}}.

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