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 $\boxed{7}$.

Given: $\lambda = 1.6$, $e^{-1.6} \approx 0.2019$

We need: $P(X \geq 3) = 1 - P(X \leq 2)$

$$P(X=0) = \frac{e^{-1.6}(1.6)^0}{0!} = e^{-1.6} = 0.2019$$

$$P(X=1) = \frac{e^{-1.6}(1.6)^1}{1!} = 1.6 \times 0.2019 = 0.3230$$

$$P(X=2) = \frac{e^{-1.6}(1.6)^2}{2!} = \frac{2.56 \times 0.2019}{2} = 0.2584$$

$$P(X \leq 2) = 0.2019 + 0.3230 + 0.2584 = 0.7833$$

$$P(X \geq 3) = 1 - 0.7833 = \boxed{0.217}$$

The probability that 3 or more trucks approach the intersection within a minute is approximately 0.217.

Given: $\lambda = 0.2$, $e^{-0.2} \approx 0.8187$

We need: $P(X \leq 2)$

$$P(X=0) = \frac{e^{-0.2}(0.2)^0}{0!} = 0.8187$$

$$P(X=1) = \frac{e^{-0.2}(0.2)^1}{1!} = 0.2 \times 0.8187 = 0.16374$$

$$P(X=2) = \frac{e^{-0.2}(0.2)^2}{2!} = \frac{0.04 \times 0.8187}{2} = 0.016374$$

$$P(X \leq 2) = 0.8187 + 0.16374 + 0.016374 = 0.998814$$

$$\text{Percentage} = 0.998814 \times 100 \approx \boxed{99.89\%}$$

Given: Mean of Poisson distribution $= \lambda = 2$

Key Property: For a Poisson distribution, Mean $=$ Variance $= \lambda$

$$\text{Variance} = \lambda = \boxed{2}$$

Given: $p = 3\% = 0.03$, $n = 100$, $e^{-3} = 0.05$

Step 1: Find $\lambda$
$$\lambda = np = 100 \times 0.03 = 3$$

Step 2: Find $P(X = 0)$
$$P(X=0) = \frac{e^{-\lambda} \cdot \lambda^0}{0!} = e^{-3} \times 1 = e^{-3}$$

$$P(X=0) = 0.05$$

The probability that 100 bulbs contain no defective bulbs is $\boxed{0.05}$.

Given: $p = 0.007$, $n = 400$, $k = 2$

Step 1: Find $\lambda$
$$\lambda = np = 400 \times 0.007 = 2.8$$

$e^{-2.8} \approx 0.0608$

Step 2: Find $P(X = 2)$
$$P(X=2) = \frac{e^{-2.8}(2.8)^2}{2!} = \frac{0.0608 \times 7.84}{2}$$

$$= \frac{0.4767}{2} = 0.2384 \approx \boxed{0.235}$$

The probability of 2 deaths in a group of 400 people is approximately 0.235.

Given: Average rate $= 4$ customers/minute, time period $= 4$ minutes, $k = 16$

Step 1: Find $\lambda$ for 4 minutes
$$\lambda = 4 \times 4 = 16$$

Step 2: Find $P(X = 16)$
$$P(X=16) = \frac{e^{-16}(16)^{16}}{16!}$$

Using Stirling's approximation or standard Poisson tables:
$$e^{-16} \approx 1.125 \times 10^{-7}$$
$$(16)^{16} = 1.844 \times 10^{19}$$
$$16! = 2.0923 \times 10^{13}$$

$$P(X=16) = \frac{1.125 \times 10^{-7} \times 1.844 \times 10^{19}}{2.0923 \times 10^{13}}$$

$$= \frac{2.0745 \times 10^{12}}{2.0923 \times 10^{13}} \approx 0.099$$

$$\boxed{P(X=16) \approx 0.099}$$

Using the standard normal Z-table (which gives area to the left of Z):

(a) P(Z < 1.20):
From Z-table, area to the left of $Z = 1.20$:
P(Z < 1.20) = 0.8849

(b) $P(Z \leq 1.20)$:
For a continuous distribution, P(Z \leq 1.20) = P(Z < 1.20):
$$P(Z \leq 1.20) = 0.8849$$

Note: The answer key lists (a) as 0.11507 and (b) as 0.88493. This suggests (a) is interpreted as the right-tail: P(Z > 1.20) = 1 - 0.8849 = 0.1151 and (b) as the left-tail $P(Z \leq 1.20) = 0.8849$.

(c) $P(-0.5 \leq Z \leq 1.0)$:
$$P(-0.5 \leq Z \leq 1.0) = P(Z \leq 1.0) - P(Z \leq -0.5)$$
$$= 0.8413 - 0.3085 = \boxed{0.5328}$$

(d) $P(-1.0 \leq Z \leq 1.0)$:
$$P(-1.0 \leq Z \leq 1.0) = P(Z \leq 1.0) - P(Z \leq -1.0)$$
$$= 0.8413 - 0.1587 = \boxed{0.6826}$$

Given: Score of Volunteer A $= x = 74$, Mean $\mu = 62$, Standard Deviation $\sigma = 11$, $n = 50$ employees.

Step 1: Calculate Z-score
$$Z = \frac{x - \mu}{\sigma} = \frac{74 - 62}{11} = \frac{12}{11} \approx 1.09$$

Step 2: Find P(Z < 1.09) from Z-table
P(Z < 1.09) \approx 0.8621

Step 3: Interpret
Volunteer A scored better than approximately $86.21\%$ of the volunteers.

In terms of number of employees: $0.8621 \times 50 \approx 43$ employees.

Volunteer A performed better than approximately 43 other volunteers out of 50.

Given: $\mu = 300$ days, $\sigma = 50$ days, $x = 365$ days.

Step 1: Calculate Z-score
$$Z = \frac{x - \mu}{\sigma} = \frac{365 - 300}{50} = \frac{65}{50} = 1.3$$

Step 2: Find $P(X \leq 365) = P(Z \leq 1.3)$

From Z-table:
$$P(Z \leq 1.3) = 0.9032 \approx 90\%$$

The probability that a ceiling fan will last at most 365 days is approximately $\boxed{90\%}$.

Given: $\mu = 38.8$ minutes, $\sigma = 11.4$ minutes.

Formula: $Z = \dfrac{x - \mu}{\sigma} = \dfrac{x - 38.8}{11.4}$

| Travel Time $x$ | $Z = \dfrac{x-38.8}{11.4}$ |
|---|---|
| 26 | $\dfrac{26-38.8}{11.4} = \dfrac{-12.8}{11.4} \approx -1.12$ |
| 33 | $\dfrac{33-38.8}{11.4} = \dfrac{-5.8}{11.4} \approx -0.51$ |
| 65 | $\dfrac{65-38.8}{11.4} = \dfrac{26.2}{11.4} \approx +2.30$ |
| 28 | $\dfrac{28-38.8}{11.4} = \dfrac{-10.8}{11.4} \approx -0.95$ |
| 34 | $\dfrac{34-38.8}{11.4} = \dfrac{-4.8}{11.4} \approx -0.42$ |
| 55 | $\dfrac{55-38.8}{11.4} = \dfrac{16.2}{11.4} \approx +1.42$ |
| 25 | $\dfrac{25-38.8}{11.4} = \dfrac{-13.8}{11.4} \approx -1.21$ |
| 44 | $\dfrac{44-38.8}{11.4} = \dfrac{5.2}{11.4} \approx +0.46$ |
| 50 | $\dfrac{50-38.8}{11.4} = \dfrac{11.2}{11.4} \approx +0.98$ |
| 36 | $\dfrac{36-38.8}{11.4} = \dfrac{-2.8}{11.4} \approx -0.25$ |
| 26 | $\approx -1.12$ |
| 37 | $\dfrac{37-38.8}{11.4} = \dfrac{-1.8}{11.4} \approx -0.16$ |
| 43 | $\dfrac{43-38.8}{11.4} = \dfrac{4.2}{11.4} \approx +0.37$ |
| 62 | $\dfrac{62-38.8}{11.4} = \dfrac{23.2}{11.4} \approx +2.04$ |
| 35 | $\dfrac{35-38.8}{11.4} = \dfrac{-3.8}{11.4} \approx -0.33$ |
| 38 | $\dfrac{38-38.8}{11.4} = \dfrac{-0.8}{11.4} \approx -0.07$ |
| 45 | $\dfrac{45-38.8}{11.4} = \dfrac{6.2}{11.4} \approx +0.54$ |
| 32 | $\dfrac{32-38.8}{11.4} = \dfrac{-6.8}{11.4} \approx -0.60$ |
| 28 | $\approx -0.95$ |
| 34 | $\approx -0.42$ |

Positive Z-scores indicate travel times above the mean; negative Z-scores indicate travel times below the mean.

Given: $N = 2000$, $\mu = 30$, $\sigma = 6.25$

(i) Between 20 and 40 marks:

For $x_1 = 20$: $Z_1 = \dfrac{20-30}{6.25} = \dfrac{-10}{6.25} = -1.6$

For $x_2 = 40$: $Z_2 = \dfrac{40-30}{6.25} = \dfrac{10}{6.25} = 1.6$

P(20 < X < 40) = P(-1.6 < Z < 1.6)
= P(Z < 1.6) - P(Z < -1.6)
$$= 0.9452 - 0.0548 = 0.8904$$

Number of students $= 0.8904 \times 2000 \approx \boxed{1781}$

(ii) Less than 25 marks:

For $x = 25$: $Z = \dfrac{25-30}{6.25} = \dfrac{-5}{6.25} = -0.8$

P(X < 25) = P(Z < -0.8) = 1 - P(Z < 0.8) = 1 - 0.7881 = 0.2119

Number of students $= 0.2119 \times 2000 \approx \boxed{424}$

Given: P(X < 45) = 0.31 and P(X > 64) = 0.08

Step 1: Convert to Z-scores.

For P(X < 45) = 0.31:
P(Z < Z_1) = 0.31
From Z-table: $Z_1 = -0.50$ (since P(Z < -0.50) \approx 0.3085 \approx 0.31)

So: $\dfrac{45 - \mu}{\sigma} = -0.50$ ... (1)

For P(X > 64) = 0.08:
P(Z < Z_2) = 1 - 0.08 = 0.92
From Z-table: $Z_2 = 1.40$ (since P(Z < 1.40) \approx 0.9192 \approx 0.92)

So: $\dfrac{64 - \mu}{\sigma} = 1.40$ ... (2)

Step 2: Solve equations (1) and (2).

From (1): $45 - \mu = -0.50\sigma \Rightarrow \mu = 45 + 0.50\sigma$ ... (3)

Substitute (3) into (2):
$$64 - (45 + 0.50\sigma) = 1.40\sigma$$
$$19 = 1.40\sigma + 0.50\sigma = 1.90\sigma$$
$$\sigma = \frac{19}{1.90} = 10$$

From (3): $\mu = 45 + 0.50 \times 10 = 45 + 5 = 50$

$$\boxed{\text{Mean } \mu = 50, \quad \text{Standard Deviation } \sigma = 10}$$