The events S and T are independent. - P(S and T) = \frac{1}{6} - P(S) = \frac{1}{4} 10.1.1 Calculate P(T). 10.1.2 Hence, calculate P(S or T). 10.2 A FIVE-digit code is created from the digits 2 ; 3 ; 5 ; 7 ; 9. How many different codes can be created if: 10.2.1 Repetition of digits is NOT allowed in the code. 10.2.2 Repetition of digits IS allowed in the code. 10.3 A group of 3 South Africans, 2 Australians and 2 Englishmen are staying at the same hotel while on holiday. Each person has his/her own room and the rooms are next to each other in a straight corridor. If the rooms are allocated at random, determine the probability that the 2 Australians will have adjacent rooms and the 2 Englishmen will also have adjacent rooms.

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The events S and T are independent - NSC Mathematics - Question 10 - 2017 - Paper 1

Question 10

The events S and T are independent. - P(S and T) = \frac{1}{6} - P(S) = \frac{1}{4} 10.1.1 Calculate P(T). 10.1.2 Hence, calculate P(S or T). 10.2 A FIVE-digit c... show full transcript

Worked Solution & Example Answer:The events S and T are independent - NSC Mathematics - Question 10 - 2017 - Paper 1

Step 1

Calculate P(T)

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Answer

To find P(T), we can use the formula for independent events:

$P(S \text{ and } T) = P(S) \times P(T)$

We know:

$\frac{1}{6} = \frac{1}{4} \times P(T)$

To find P(T), rearranging gives us:

$P(T) = \frac{1/6}{1/4} = \frac{1}{6} \times \frac{4}{1} = \frac{4}{6} = \frac{2}{3}$

Step 2

Hence, calculate P(S or T)

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Answer

To find P(S or T), we can use the formula:

$P(S \text{ or } T) = P(S) + P(T) - P(S \text{ and } T)$

Substituting the known values:

$P(S) = \frac{1}{4}, P(T) = \frac{2}{3}$

Now we substitute and calculate:

$P(S \text{ or } T) = \frac{1}{4} + \frac{2}{3} - \frac{1}{6}$

Finding a common denominator (12):

$= \frac{3}{12} + \frac{8}{12} - \frac{2}{12} = \frac{9}{12} = \frac{3}{4}$

Step 3

10.2.1 Repetition of digits is NOT allowed in the code

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Answer

Since repetition is not allowed, we can choose any 5 digits from the set of {2, 3, 5, 7, 9}. The total number of arrangements with 5 distinct digits is:

$5! = 120$

Step 4

10.2.2 Repetition of digits IS allowed in the code

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Answer

If repetition is allowed, each digit can be any of the 5 digits for each of the 5 places. Therefore, the total number of codes is:

$5^5 = 3125$

Step 5

10.3 Determine the probability of adjacent rooms

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Answer

First, we will consider the two Australians as a single unit (block), so we have:

Total units = 3 South Africans + 1 Australian block + 2 Englishmen = 6 units.

The arrangements of these units: $6!$

The Australians themselves can switch places: $2!$

So the total arrangements = $6! * 2!$

Now, for the two Englishmen, treating them as another block gives:

Total units = 5

Total arrangements = $5! * 2!$

Thus, the probability:

$P = \frac{6! * 2! * 5! * 2!}{(7!)}$

Calculating gives:

$= \frac{720 * 2 * 120 * 2}{5040} = \frac{34560}{5040} = \frac{2}{21}$

The events S and T are independent - NSC Mathematics - Question 10 - 2017 - Paper 1

Worked Solution & Example Answer:The events S and T are independent - NSC Mathematics - Question 10 - 2017 - Paper 1

Calculate P(T)

Hence, calculate P(S or T)

10.2.1 Repetition of digits is NOT allowed in the code

10.2.2 Repetition of digits IS allowed in the code

10.3 Determine the probability of adjacent rooms

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