Sure, let's tackle each part of the question step by step.
### Sample Space
When two coins are tossed simultaneously, there are four possible outcomes:
1. Both coins show Heads (HH)
2. The first coin shows Heads and the second shows Tails (HT)
3. The first coin shows Tails and the second shows Heads (TH)
4. Both coins show Tails (TT)
So, the sample space is: {HH, HT, TH, TT}
There are 4 outcomes in the sample space.
### (a) P(2 tails)
To find the probability of getting 2 tails, we look for outcomes where both coins show Tails.
- The favorable outcome is: {TT}
There is 1 such favorable outcome.
So, the probability [tex]\(P(\text{2 tails}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{4} = 0.25\)[/tex].
### (b) P(exactly one tail)
To find the probability of getting exactly one tail, we look for outcomes where one coin shows Heads, and the other shows Tails.
- The favorable outcomes are: {HT, TH}
There are 2 such favorable outcomes.
So, the probability [tex]\(P(\text{exactly one tail}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{4} = 0.5\)[/tex].
### (c) P(no tails)
To find the probability of getting no tails, we look for outcomes where both coins show Heads.
- The favorable outcome is: {HH}
There is 1 such favorable outcome.
So, the probability [tex]\(P(\text{no tails}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{4} = 0.25\)[/tex].
### (d) P(at most one head)
To find the probability of getting at most one head, we look for outcomes where there is zero or one head.
- The favorable outcomes are: {TT, HT, TH}
There are 3 such favorable outcomes.
So, the probability [tex]\(P(\text{at most one head}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{4} = 0.75\)[/tex].
### (e) P(one head)
To find the probability of getting exactly one head, we look for outcomes where one coin shows Heads, and the other shows Tails (similar to part b).
- The favorable outcomes are: {HT, TH}
There are 2 such favorable outcomes.
So, the probability [tex]\(P(\text{one head}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{4} = 0.5\)[/tex].
Thus, the step-by-step probabilities for each part of the question are:
(a) [tex]\(P(\text{2 tails}) = 0.25\)[/tex]
(b) [tex]\(P(\text{exactly one tail}) = 0.5\)[/tex]
(c) [tex]\(P(\text{no tails}) = 0.25\)[/tex]
(d) [tex]\(P(\text{at most one head}) = 0.75\)[/tex]
(e) [tex]\(P(\text{one head}) = 0.5\)[/tex]
