22. A lorry load of potatoes has, on average, one rotten potato in every 6. A greengrocer tests a random sample of 100 potatoes and decides to turn away the lorry if she finds more than 18 rotten potatoes in the sample. Find the probability that she accepts the consignment.

23. On average, 1 out of 10 game consoles purchased is spoilt.

(a) If 20 game consoles were bought, what is the probability that:
(i) More than 17 game consoles are not spoilt?
(ii) At least one is spoilt?

(b) If 400 game consoles were bought, find the probability that 50 to 70 game consoles are spoilt.

The problem here involves a binomial distribution. Given that on average, one in six potatoes is rotten, the probability [tex]p[/tex] of picking a rotten potato is [tex]p = \frac{1}{6}[/tex].
The greengrocer's sample size [tex]n[/tex] is 100 potatoes.
The greengrocer will accept the consignment if she finds 18 or fewer rotten potatoes out of 100 sampled, which means we need to find [tex]P(X \leq 18)[/tex].

We'll use the binomial probability formula:

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

Since calculating [tex]P(X \leq 18)[/tex] directly with the binomial distribution is complex, it is often practical to use a normal approximation when [tex]n[/tex] is large.

The mean [tex]\mu[/tex] of the distribution is given by [tex]\mu = np = 100 \times \frac{1}{6} = 16.67[/tex].
The standard deviation [tex]\sigma[/tex] is [tex]\sigma = \sqrt{np(1-p)} = \sqrt{100 \times \frac{1}{6} \times \frac{5}{6}} \approx 3.61[/tex].

Using the normal approximation, find the z-score for 18:

[tex]z = \frac{18 + 0.5 - 16.67}{3.61} \approx 0.51[/tex]

Looking up this z-score in a standard normal distribution table, we find:

[tex]P(X \leq 18) \approx 0.695[/tex]

So, there is approximately a 69.5% chance that the greengrocer will accept the consignment.

23. Probability Questions on Spoilt Game Consoles

(a) Given: 1 out of 10 consoles is spoilt. Therefore, [tex]p = \frac{1}{10}[/tex] and [tex]1-p = \frac{9}{10}[/tex].

(i) Probability more than 17 are not spoilt:

Find the probability that 18, 19, or 20 consoles are not spoilt out of 20.

Using the binomial formula:

[tex]P(X > 17) = P(X = 18) + P(X = 19) + P(X = 20)[/tex]

Calculate each:

[tex]P(X = 18) = \binom{20}{18} \left(\frac{9}{10}\right)^{18} \left(\frac{1}{10}\right)^{2} \approx 0.2852[/tex]
[tex]P(X = 19) = \binom{20}{19} \left(\frac{9}{10}\right)^{19} \left(\frac{1}{10}\right)^{1} \approx 0.3774[/tex]
[tex]P(X = 20) = \binom{20}{20} \left(\frac{9}{10}\right)^{20} = \left(\frac{9}{10}\right)^{20} \approx 0.1216[/tex]

Summing these probabilities:

[tex]P(X > 17) \approx 0.2852 + 0.3774 + 0.1216 \approx 0.7842[/tex]

(ii) Probability at least one is spoilt:

Easier to calculate by finding the probability none are spoilt and subtracting from 1.

[tex]P( ext{none spoilt}) = \left(\frac{9}{10}\right)^{20} \approx 0.1216[/tex]

So,

[tex]P( ext{at least one spoilt}) = 1 - P( ext{none spoilt}) \approx 1 - 0.1216 = 0.8784[/tex]

(b) Probability that 50 to 70 are spoilt out of 400:

Here, [tex]n = 400[/tex], [tex]p = \frac{1}{10}[/tex], use normal approximation.

Calculate mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex]:

[tex]\mu = np = 400 \times \frac{1}{10} = 40[/tex]
[tex]\sigma = \sqrt{np(1-p)} = \sqrt{400 \times \frac{1}{10} \times \frac{9}{10}} \approx 6[/tex]

Find the z-scores for 50 and 70:

[tex]z_{50} = \frac{50 + 0.5 - 40}{6} \approx 1.75[/tex]
[tex]z_{70} = \frac{70 + 0.5 - 40}{6} \approx 5.08[/tex]

Using a standard normal distribution table, it is evident that [tex]P(50 < X < 70) \approx P(z_{50} < Z < z_{70}) \approx P(Z < 1.75) - P(Z < 5.08)[/tex].

Since [tex]P(Z < 5.08) \approx 1[/tex],

[tex]P(z_{50} < Z < z_{70}) \approx 0.9599[/tex]).

Thus, the approximate probability is 95.9%.