Probability Distributions

For a valid probability distribution two conditions must hold:
(i) Each $P(X) \geq 0$
(ii) $\sum P(X) = 1$

(a)
$$\sum P(X) = 0.1 + 0.8 + 0.001 + 0.2 = 1.101 \neq 1$$
Since the sum of probabilities is not equal to 1, this is NOT a valid probability distribution.

(b)
$P(X = 4) = -0.2 < 0$
Since one of the probabilities is negative, this is NOT a valid probability distribution.

(c)
$$\sum P(X) = 0.5 + 0.2 + 0.3 = 1.0$$
All probabilities are non-negative and their sum equals 1.
Hence, this IS a valid probability distribution.

Given: 2 black pens and 1 red pen. Drawing is done with replacement, twice.

Probability of drawing a red pen in one draw: $p = \dfrac{1}{3}$

Probability of drawing a black pen in one draw: $q = \dfrac{2}{3}$

X = number of red pens in two draws, so $X$ can take values $0, 1, 2$.

$$P(X=0) = \binom{2}{0}\left(\frac{1}{3}\right)^0\left(\frac{2}{3}\right)^2 = \frac{4}{9}$$

$$P(X=1) = \binom{2}{1}\left(\frac{1}{3}\right)^1\left(\frac{2}{3}\right)^1 = \frac{4}{9}$$

$$P(X=2) = \binom{2}{2}\left(\frac{1}{3}\right)^2\left(\frac{2}{3}\right)^0 = \frac{1}{9}$$

Probability Distribution Table:

| $X$ | 0 | 1 | 2 |
|---|---|---|---|
| $P(X)$ | $\dfrac{4}{9}$ | $\dfrac{4}{9}$ | $\dfrac{1}{9}$ |

Verification: $\dfrac{4}{9}+\dfrac{4}{9}+\dfrac{1}{9} = 1$ ✓

Given: A die with 1 on 3 faces, 2 on 2 faces, 5 on 1 face.

The probability distribution is:

| $X$ | 1 | 2 | 5 |
|---|---|---|---|
| $P(X)$ | $\dfrac{3}{6}=\dfrac{1}{2}$ | $\dfrac{2}{6}=\dfrac{1}{3}$ | $\dfrac{1}{6}$ |

Mean $= E(X) = \sum x_i P(x_i)$

$$E(X) = 1 \times \frac{1}{2} + 2 \times \frac{1}{3} + 5 \times \frac{1}{6}$$

$$= \frac{1}{2} + \frac{2}{3} + \frac{5}{6}$$

$$= \frac{3}{6} + \frac{4}{6} + \frac{5}{6} = \frac{12}{6} = \boxed{2}$$

The mean of the numbers obtained is 2.

Given: $n = 10$, fair coin so $p = \dfrac{1}{2}$, $q = \dfrac{1}{2}$.

Using Binomial Distribution: $P(X=r) = \binom{n}{r}p^r q^{n-r}$

**(i) Exactly six heads ($r = 6$):**
$$P(X=6) = \binom{10}{6}\left(\frac{1}{2}\right)^6\left(\frac{1}{2}\right)^4 = \binom{10}{6}\left(\frac{1}{2}\right)^{10}$$
$$= 210 \times \frac{1}{1024} = \frac{210}{1024} = \frac{105}{512}$$

**(ii) At least six heads ($r \geq 6$):**
$$P(X \geq 6) = P(6)+P(7)+P(8)+P(9)+P(10)$$
$$= \frac{1}{1024}\left[\binom{10}{6}+\binom{10}{7}+\binom{10}{8}+\binom{10}{9}+\binom{10}{10}\right]$$
$$= \frac{1}{1024}\left[210+120+45+10+1\right] = \frac{386}{1024} = \frac{193}{512}$$

**(iii) At most six heads ($r \leq 6$):**
$$P(X \leq 6) = 1 - P(X \geq 7)$$
$$P(X \geq 7) = \frac{1}{1024}[120+45+10+1] = \frac{176}{1024} = \frac{11}{64}$$
$$P(X \leq 6) = 1 - \frac{11}{64} = \frac{53}{64}$$

Given: $n = 7$ throws, success = getting a 5.

$p = P(\text{getting 5}) = \dfrac{1}{6}$, $q = 1 - \dfrac{1}{6} = \dfrac{5}{6}$, $r = 2$

Using Binomial Distribution:
$$P(X=2) = \binom{7}{2}\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^5$$

$$= 21 \times \frac{1}{36} \times \frac{5^5}{6^5}$$

$$= \frac{21 \times 3125}{36 \times 7776} = \frac{21 \times 5^5}{6^7}$$

$$= \binom{7}{2}\frac{5^5}{6^7}$$

$$\boxed{P(X=2) = \frac{21 \times 3125}{279936} \approx 0.2344}$$

Given: The probability distribution with unknown constant $k$.

(a) Finding k:

Since $\sum P(X=x_i) = 1$:
$$P(0)+P(1)+P(2)+P(3)+P(4) = 1$$
$$0.1 + k(1) + k(2) + k(5-3) + k(5-4) = 1$$
$$0.1 + k + 2k + 2k + k = 1$$
$$0.1 + 6k = 1$$
$$6k = 0.9$$
$$\boxed{k = 0.15}$$

The probability distribution becomes:

| $x_i$ | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| $P(X)$ | 0.1 | 0.15 | 0.30 | 0.30 | 0.15 |

(b)

At least two hours ($X \geq 2$):
$$P(X \geq 2) = P(2)+P(3)+P(4) = 0.30+0.30+0.15 = \boxed{0.75}$$

Exactly two hours ($X = 2$):
$$P(X=2) = k \times 2 = 0.15 \times 2 = \boxed{0.30}$$

At most two hours ($X \leq 2$):
$$P(X \leq 2) = P(0)+P(1)+P(2) = 0.1+0.15+0.30 = \boxed{0.55}$$

Given: Fair coin, $p = \dfrac{1}{2}$, $q = \dfrac{1}{2}$.

We need: $P(\text{at least one head}) > 0.90$

$$P(\text{at least one head}) = 1 - P(\text{no heads}) = 1 - \left(\frac{1}{2}\right)^n > 0.90$$

$$\left(\frac{1}{2}\right)^n < 0.10$$

Checking values:
- $n=1$: $(1/2)^1 = 0.5 > 0.10$ ✗
- $n=2$: $(1/2)^2 = 0.25 > 0.10$ ✗
- $n=3$: $(1/2)^3 = 0.125 > 0.10$ ✗
- $n=4$: $(1/2)^4 = 0.0625 < 0.10$ ✓

For $n = 4$: $P(\text{at least one head}) = 1 - 0.0625 = 0.9375 > 90\%$ ✓

Sumit must toss the coin at least 4 times.

Given: Two dice are thrown. $X$ = sum of numbers on two dice.

Total outcomes = 36.

The possible values of $X$ range from 2 to 12.

| Sum $X$ | Frequency | $P(X)$ | $X \cdot P(X)$ |
|---|---|---|---|
| 2 | 1 | 1/36 | 2/36 |
| 3 | 2 | 2/36 | 6/36 |
| 4 | 3 | 3/36 | 12/36 |
| 5 | 4 | 4/36 | 20/36 |
| 6 | 5 | 5/36 | 30/36 |
| 7 | 6 | 6/36 | 42/36 |
| 8 | 5 | 5/36 | 40/36 |
| 9 | 4 | 4/36 | 36/36 |
| 10 | 3 | 3/36 | 30/36 |
| 11 | 2 | 2/36 | 22/36 |
| 12 | 1 | 1/36 | 12/36 |

$$E(X) = \sum X \cdot P(X) = \frac{2+6+12+20+30+42+40+36+30+22+12}{36}$$

$$= \frac{252}{36} = \boxed{7}$$

The mathematical expectation of $X$ is 7.

Given: $p = 0.6$ (probability of success)

$q = 1 - p = 1 - 0.6 = 0.4$

For a Bernoulli distribution, $n = 1$:
$$\text{Variance} = npq = 1 \times p \times q = p \times q$$

$$\text{Var}(X) = 0.6 \times 0.4 = \boxed{0.24}$$

Given: Mean $= np = \dfrac{4}{3}$, Variance $= npq = \dfrac{8}{9}$

Step 1: Find q
$$q = \frac{\text{Variance}}{\text{Mean}} = \frac{8/9}{4/3} = \frac{8}{9} \times \frac{3}{4} = \frac{2}{3}$$

Step 2: Find p
$$p = 1 - q = 1 - \frac{2}{3} = \frac{1}{3}$$

Step 3: Find n
$$np = \frac{4}{3} \Rightarrow n \times \frac{1}{3} = \frac{4}{3} \Rightarrow n = 4$$

**Step 4: Find $P(X=1)$**
$$P(X=1) = \binom{4}{1}\left(\frac{1}{3}\right)^1\left(\frac{2}{3}\right)^3$$

$$= 4 \times \frac{1}{3} \times \frac{8}{27} = \frac{32}{81}$$

$$\boxed{P(X=1) = \frac{32}{81}}$$

Given: A fair coin is tossed once. Let $X$ = number of tails.

$X$ can take values 0 or 1.

$P(X=0) = \dfrac{1}{2}$ (head), $P(X=1) = \dfrac{1}{2}$ (tail)

$$E(X) = 0 \times \frac{1}{2} + 1 \times \frac{1}{2} = \frac{1}{2} = \boxed{0.5}$$

The expected number of tails is 0.5.

Given: $\lambda = 4.5$ calls per 5 minutes. Calls follow Poisson distribution.

We need to find minimum $n$ (number of executives) such that:
$$P(X > n) \leq 0.10 \quad \Rightarrow \quad P(X \leq n) \geq 0.90$$

Using Poisson formula: $P(X=k) = \dfrac{e^{-4.5}(4.5)^k}{k!}$, where $e^{-4.5} \approx 0.01111$

Computing cumulative probabilities:

| $k$ | $P(X=k)$ | $P(X \leq k)$ |
|---|---|---|
| 0 | 0.01111 | 0.01111 |
| 1 | 0.05000 | 0.06111 |
| 2 | 0.11248 | 0.17359 |
| 3 | 0.16873 | 0.34232 |
| 4 | 0.18982 | 0.53214 |
| 5 | 0.17084 | 0.70298 |
| 6 | 0.12813 | 0.83111 |
| 7 | 0.08237 | 0.91348 |

At $n = 7$: $P(X \leq 7) \approx 0.9135 \geq 0.90$ ✓

At $n = 6$: $P(X \leq 6) \approx 0.831 < 0.90$ ✗

**The minimum number of customer executives needed is $\boxed{7}$.**