(i) A coin is tossed 5 times. What is the probability of getting :
(a) at least 3 heads (b) at most 2 heads
( c ) no head (d) 3 heads ?
(ii) If a fair coin is tossed 10 times, find the probability of :
(a) exactly four heads
(b) exactly six heads
( c ) at least six heads
(d) at most six heads.

Answer

To solve the problem step by step, we will use the binomial probability formula, which is given by:

\[ P(X = r) = \binom{n}{r} p^r (1-p)^{n-r} \]

where:
- \( n \) = number of trials (coin tosses)
- \( r \) = number of successful outcomes (heads)
- \( p \) = probability of success on an individual trial (for a fair coin, \( p = \frac{1}{2} \))
- \( \binom{n}{r} \) = binomial coefficient, calculated as \( \frac{n!}{r!(n-r)!} \)

### Part (i): A coin is tossed 5 times

#### (a) Probability of getting at least 3 heads

1. Calculate \( P(X \geq 3) \):
\[ P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5) \]

2. Calculate each probability:
- \( P(X = 3) \):
\[ P(X = 3) = \binom{5}{3} \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^{5-3} = \binom{5}{3} \left(\frac{1}{2}\right)^5 = 10 \cdot \frac{1}{32} = \frac{10}{32} \]

- \( P(X = 4) \):
\[ P(X = 4) = \binom{5}{4} \left(\frac{1}{2}\right)^4 \left(\frac{1}{2}\right)^{5-4} = \binom{5}{4} \left(\frac{1}{2}\right)^5 = 5 \cdot \frac{1}{32} = \frac{5}{32} \]

- \( P(X = 5) \):
\[ P(X = 5) = \binom{5}{5} \left(\frac{1}{2}\right)^5 \left(\frac{1}{2}\right)^{5-5} = \binom{5}{5} \left(\frac{1}{2}\right)^5 = 1 \cdot \frac{1}{32} = \frac{1}{32} \]

3. Sum the probabilities:
\[ P(X \geq 3) = \frac{10}{32} + \frac{5}{32} + \frac{1}{32} = \frac{16}{32} = \frac{1}{2} \]

#### (b) Probability of getting at most 2 heads

1. Calculate \( P(X \leq 2) \):
\[ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \]

2. Calculate each probability:
- \( P(X = 0) \):
\[ P(X = 0) = \binom{5}{0} \left(\frac{1}{2}\right)^0 \left(\frac{1}{2}\right)^{5} = 1 \cdot \frac{1}{32} = \frac{1}{32} \]

- \( P(X = 1) \):
\[ P(X = 1) = \binom{5}{1} \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^{4} = 5 \cdot \frac{1}{32} = \frac{5}{32} \]

- \( P(X = 2) \):
\[ P(X = 2) = \binom{5}{2} \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^{3} = 10 \cdot \frac{1}{32} = \frac{10}{32} \]

3. Sum the probabilities:
\[ P(X \leq 2) = \frac{1}{32} + \frac{5}{32} + \frac{10}{32} = \frac{16}{32} = \frac{1}{2} \]

#### (c) Probability of getting no heads

1. Calculate \( P(X = 0) \):
\[ P(X = 0) = \binom{5}{0} \left(\frac{1}{2}\right)^0 \left(\frac{1}{2}\right)^{5} = 1 \cdot \frac{1}{32} = \frac{1}{32} \]

#### (d) Probability of getting exactly 3 heads

1. Calculate \( P(X = 3) \):
\[ P(X = 3) = \binom{5}{3} \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^{2} = 10 \cdot \frac{1}{32} = \frac{10}{32} = \frac{5}{16} \]

### Part (ii): A fair coin is tossed 10 times

#### (a) Probability of exactly four heads

1. Calculate \( P(X = 4) \):
\[ P(X = 4) = \binom{10}{4} \left(\frac{1}{2}\right)^{4} \left(\frac{1}{2}\right)^{6} = \binom{10}{4} \left(\frac{1}{2}\right)^{10} \]
\[ \binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210 \]
\[ P(X = 4) = 210 \cdot \frac{1}{1024} = \frac{210}{1024} = \frac{105}{512} \]

#### (b) Probability of exactly six heads

1. Calculate \( P(X = 6) \):
\[ P(X = 6) = \binom{10}{6} \left(\frac{1}{2}\right)^{6} \left(\frac{1}{2}\right)^{4} = \binom{10}{6} \left(\frac{1}{2}\right)^{10} \]
\[ \binom{10}{6} = \binom{10}{4} = 210 \]
\[ P(X = 6) = 210 \cdot \frac{1}{1024} = \frac{210}{1024} = \frac{105}{512} \]

#### (c) Probability of at least six heads

1. Calculate \( P(X \geq 6) \):
\[ P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) \]

2. Calculate each probability:
- \( P(X = 7) \):
\[ P(X = 7) = \binom{10}{7} \left(\frac{1}{2}\right)^{10} = 120 \cdot \frac{1}{1024} = \frac{120}{1024} = \frac{15}{128} \]

- \( P(X = 8) \):
\[ P(X = 8) = \binom{10}{8} \left(\frac{1}{2}\right)^{10} = 45 \cdot \frac{1}{1024} = \frac{45}{1024} \]

- \( P(X = 9) \):
\[ P(X = 9) = \binom{10}{9} \left(\frac{1}{2}\right)^{10} = 10 \cdot \frac{1}{1024} = \frac{10}{1024} = \frac{5}{512} \]

- \( P(X = 10) \):
\[ P(X = 10) = \binom{10}{10} \left(\frac{1}{2}\right)^{10} = 1 \cdot \frac{1}{1024} = \frac{1}{1024} \]

3. Sum the probabilities:
\[ P(X \geq 6) = \frac{105}{512} + \frac{15}{128} + \frac{45}{1024} + \frac{5}{512} + \frac{1}{1024} \]

Convert to a common denominator (1024):
\[ P(X \geq 6) = \frac{210}{1024} + \frac{120}{1024} + \frac{45}{1024} + \frac{10}{1024} + \frac{1}{1024} = \frac{386}{1024} = \frac{193}{512} \]

#### (d) Probability of at most six heads

1. Calculate \( P(X \leq 6) \):
\[ P(X \leq 6) = 1 - P(X \geq 7) \]
\[ P(X \leq 6) = 1 - \left(P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)\right) \]
\[ P(X \leq 6) = 1 - \left(\frac{120}{1024} + \frac{45}{1024} + \frac{10}{1024} + \frac{1}{1024}\right) \]
\[ P(X \leq 6) = 1 - \frac{176}{1024} = \frac{848}{1024} = \frac{106}{128} = \frac{53}{64} \]

### Summary of Results:
1. (i) (a) \( P(X \geq 3) = \frac{1}{2} \)
2. (i) (b) \( P(X \leq 2) = \frac{1}{2} \)
3. (i) (c) \( P(X = 0) = \frac{1}{32} \)
4. (i) (d) \( P(X = 3) = \frac{5}{16} \)
5. (ii) (a) \( P(X = 4) = \frac{105}{512} \)
6. (ii) (b) \( P(X = 6) = \frac{105}{512} \)
7. (ii) (c) \( P(X \geq 6) = \frac{193}{512} \)
8. (ii) (d) \( P(X \leq 6) = \frac{53}{64} \)