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11.1 The letters of the word EQUATION are randomly used to form a new word consisting of five letters - NSC Mathematics - Question 11 - 2017 - Paper 1

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11.1 The letters of the word EQUATION are randomly used to form a new word consisting of five letters. How many of these words are possible if letters may not be rep... show full transcript

Worked Solution & Example Answer:11.1 The letters of the word EQUATION are randomly used to form a new word consisting of five letters - NSC Mathematics - Question 11 - 2017 - Paper 1

Step 1

How many of these words are possible if letters may not be repeated?

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Answer

To determine the number of ways to arrange 5 letters from the word 'EQUATION' which has 8 distinct letters, we use the permutation formula:

P(n,r)=n!(nr)!P(n, r) = \frac{n!}{(n - r)!}

In this case, we calculate:

P(8,5)=8!(85)!=8!3!=8×7×6×5×41=6720P(8, 5) = \frac{8!}{(8 - 5)!} = \frac{8!}{3!} = \frac{8 \times 7 \times 6 \times 5 \times 4}{1} = 6720

Thus, the answer is 6720.

Step 2

Calculate P(A or B).

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Answer

Given that events A and B are independent, we can calculate P(A or B) using the formula:

P(AorB)=P(A)+P(B)P(AandB)P(A or B) = P(A) + P(B) - P(A and B)

Since P(A) and P(B) are independent, we have:

P(AandB)=P(A)×P(B)=25×0.35=0.14P(A and B) = P(A) \times P(B) = \frac{2}{5} \times 0.35 = 0.14

Now substituting the values:

P(AorB)=25+0.350.14P(A or B) = \frac{2}{5} + 0.35 - 0.14

Calculating further:

P(AorB)=0.4+0.350.14=0.61P(A or B) = 0.4 + 0.35 - 0.14 = 0.61

Therefore, P(A or B) = 0.61.

Step 3

Did not attend School A

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Answer

To find the probability that a learner did not attend School A, we consider the proportions:

Since 20% attend School A, the probability that a learner did not attend School A is:

P(notA)=1P(A)=10.2=0.8P(not A) = 1 - P(A) = 1 - 0.2 = 0.8

Thus, the probability is 0.8.

Step 4

Attended School B and failed Grade 12 in 2016

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Answer

From the data, we see that the percentage of learners attending School B is 30% with a pass rate of 50%. Thus, the failure rate is:

0.5 (failure rate)=10.5=0.50.5\text{ (failure rate)} = 1 - 0.5 = 0.5

Now, the joint probability is:

P(Bandfail)=P(B)×P(failB)=0.3×0.5=0.15P(B and fail) = P(B) \times P(fail | B) = 0.3 \times 0.5 = 0.15

Hence, the probability is 0.15.

Step 5

Passed Grade 12 in 2016

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Answer

To find the probability that a learner passed Grade 12, we will consider the pass rates from all three schools, weighted by their respective attendance:

Using the law of total probability:

P(pass)=P(A)×P(passA)+P(B)×P(passB)+P(C)×P(passC)P(pass) = P(A) \times P(pass | A) + P(B) \times P(pass | B) + P(C) \times P(pass | C)

Substituting the respective values:

P(pass)=0.2×0.8+0.3×0.5+0.5×0.9P(pass) = 0.2 \times 0.8 + 0.3 \times 0.5 + 0.5 \times 0.9

Calculating:

P(pass)=0.16+0.15+0.45=0.76P(pass) = 0.16 + 0.15 + 0.45 = 0.76

Thus, the probability that a learner passed Grade 12 in 2016 is 0.76.

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