To solve the problem step by step, we will find the probability distributions for the given scenarios.
### Part (a)
#### (i) Probability Distribution of the Number of Heads when Three Coins are Tossed
1. **Total Outcomes**: When three coins are tossed, the total number of outcomes is \(2^3 = 8\).
2. **Possible Outcomes**: The possible outcomes are:
- HHH (3 heads)
- HHT (2 heads, 1 tail)
- HTH (2 heads, 1 tail)
- THH (2 heads, 1 tail)
- HTT (1 head, 2 tails)
- THT (1 head, 2 tails)
- TTH (1 head, 2 tails)
- TTT (0 heads)
3. **Counting Heads**:
- 0 heads: 1 outcome (TTT)
- 1 head: 3 outcomes (HTT, THT, TTH)
- 2 heads: 3 outcomes (HHT, HTH, THH)
- 3 heads: 1 outcome (HHH)
4. **Probability Calculation**:
- \(P(X=0) = \frac{1}{8}\)
- \(P(X=1) = \frac{3}{8}\)
- \(P(X=2) = \frac{3}{8}\)
- \(P(X=3) = \frac{1}{8}\)
**Probability Distribution**:
- \(X = 0\), \(P(X=0) = \frac{1}{8}\)
- \(X = 1\), \(P(X=1) = \frac{3}{8}\)
- \(X = 2\), \(P(X=2) = \frac{3}{8}\)
- \(X = 3\), \(P(X=3) = \frac{1}{8}\)
#### (ii) Probability Distribution of the Number of Tails in the Simultaneous Tosses of Three Coins
The distribution of tails is complementary to that of heads. Thus, we can directly derive it:
- 0 tails (3 heads): \(P(Y=0) = \frac{1}{8}\)
- 1 tail (2 heads): \(P(Y=1) = \frac{3}{8}\)
- 2 tails (1 head): \(P(Y=2) = \frac{3}{8}\)
- 3 tails (0 heads): \(P(Y=3) = \frac{1}{8}\)
**Probability Distribution**:
- \(Y = 0\), \(P(Y=0) = \frac{1}{8}\)
- \(Y = 1\), \(P(Y=1) = \frac{3}{8}\)
- \(Y = 2\), \(P(Y=2) = \frac{3}{8}\)
- \(Y = 3\), \(P(Y=3) = \frac{1}{8}\)
#### (iii) Probability Distribution of the Number of Heads in the Simultaneous Toss of Four Coins
1. **Total Outcomes**: When four coins are tossed, the total number of outcomes is \(2^4 = 16\).
2. **Counting Heads**:
- 0 heads: 1 outcome (TTTT)
- 1 head: 4 outcomes (HTTT, THTT, TTHT, TTTT)
- 2 heads: 6 outcomes (HHTT, HTHT, HTTH, THHT, THTH, TTHH)
- 3 heads: 4 outcomes (HHHT, HHTH, HTHH, THHH)
- 4 heads: 1 outcome (HHHH)
3. **Probability Calculation**:
- \(P(X=0) = \frac{1}{16}\)
- \(P(X=1) = \frac{4}{16} = \frac{1}{4}\)
- \(P(X=2) = \frac{6}{16} = \frac{3}{8}\)
- \(P(X=3) = \frac{4}{16} = \frac{1}{4}\)
- \(P(X=4) = \frac{1}{16}\)
**Probability Distribution**:
- \(X = 0\), \(P(X=0) = \frac{1}{16}\)
- \(X = 1\), \(P(X=1) = \frac{1}{4}\)
- \(X = 2\), \(P(X=2) = \frac{3}{8}\)
- \(X = 3\), \(P(X=3) = \frac{1}{4}\)
- \(X = 4\), \(P(X=4) = \frac{1}{16}\)
### Part (b)
#### (i) Probability Distribution of the Number of Heads in Three Tosses of a Coin
This is the same as part (a)(i), so the distribution is:
- \(X = 0\), \(P(X=0) = \frac{1}{8}\)
- \(X = 1\), \(P(X=1) = \frac{3}{8}\)
- \(X = 2\), \(P(X=2) = \frac{3}{8}\)
- \(X = 3\), \(P(X=3) = \frac{1}{8}\)
#### (ii) Probability Distribution of the Number of Sixes in Two Tosses of a Die
1. **Total Outcomes**: When two dice are tossed, the total number of outcomes is \(6 \times 6 = 36\).
2. **Counting Sixes**:
- 0 sixes: 25 outcomes (not rolling a 6)
- 1 six: 10 outcomes (6 on one die, any number on the other)
- 2 sixes: 1 outcome (6 on both dice)
3. **Probability Calculation**:
- \(P(Z=0) = \frac{25}{36}\)
- \(P(Z=1) = \frac{10}{36} = \frac{5}{18}\)
- \(P(Z=2) = \frac{1}{36}\)
**Probability Distribution**:
- \(Z = 0\), \(P(Z=0) = \frac{25}{36}\)
- \(Z = 1\), \(P(Z=1) = \frac{5}{18}\)
- \(Z = 2\), \(P(Z=2) = \frac{1}{36}\)
### Part (c)
#### Probability Distribution of the Number of Heads (or Tails) in Four Tosses of a Coin
This is the same as part (a)(iii) for heads, and since the distribution of tails will be complementary, we can derive it similarly:
**Probability Distribution for Heads**:
- \(X = 0\), \(P(X=0) = \frac{1}{16}\)
- \(X = 1\), \(P(X=1) = \frac{1}{4}\)
- \(X = 2\), \(P(X=2) = \frac{3}{8}\)
- \(X = 3\), \(P(X=3) = \frac{1}{4}\)
- \(X = 4\), \(P(X=4) = \frac{1}{16}\)
**Probability Distribution for Tails**:
- \(Y = 0\), \(P(Y=0) = \frac{1}{16}\)
- \(Y = 1\), \(P(Y=1) = \frac{1}{4}\)
- \(Y = 2\), \(P(Y=2) = \frac{3}{8}\)
- \(Y = 3\), \(P(Y=3) = \frac{1}{4}\)
- \(Y = 4\), \(P(Y=4) = \frac{1}{16}\)
