Two coins are tossed. Assume that each event is equally likely to occur.
a) Use the counting principle to determine the number of sample points in the sample space.
b) Construct a tree diagram and list the sample space.
c) Determine the probability that no tails are tossed.
d) Determine the probability that exactly one tail is tossed.
e) Determine the probability that two tails are tossed.
f) Determine the probability that at least one tail is tossed.

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**a)**  
Each coin has 2 outcomes (Heads or Tails), so by the counting principle, the number of sample points is  
$$
2 \times 2 = 4.
$$

**b)**  
We can construct a tree diagram as follows:

1. First coin toss:  
 - Heads (H)  
 - Tails (T)

2. Second coin toss for each branch:
 - If first toss is H: second toss can be H or T → outcomes: HH, HT.  
 - If first toss is T: second toss can be H or T → outcomes: TH, TT.

Thus, the sample space is:  
$$
\{HH,\ HT,\ TH,\ TT\}.
$$

**c)**  
No tails tossed means both coins show Heads. There is 1 favorable outcome (HH) out of 4. Therefore, the probability is  
$$
P(\text{no tails}) = \frac{1}{4}.
$$

**d)**  
Exactly one tail tossed corresponds to the outcomes HT and TH. There are 2 favorable outcomes. Thus, the probability is  
$$
P(\text{exactly one tail}) = \frac{2}{4} = \frac{1}{2}.
$$

**e)**  
Two tails tossed corresponds to the outcome TT only. There is 1 favorable outcome. Thus, the probability is  
$$
P(\text{two tails}) = \frac{1}{4}.
$$

**f)**  
At least one tail tossed means we include outcomes with one tail or two tails: HT, TH, TT. There are 3 favorable outcomes. Thus, the probability is  
$$
P(\text{at least one tail}) = \frac{3}{4}.
$$
**a)** There are 4 possible outcomes when two coins are tossed. **b)** The sample space is: {HH, HT, TH, TT}. **c)** The probability of no tails is 1/4. **d)** The probability of exactly one tail is 1/2. **e)** The probability of two tails is 1/4. **f)** The probability of at least one tail is 3/4.

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