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11.1 You have to select a new password for your "Dropbox" account on your computer - NSC Mathematics - Question 11 - 2017 - Paper 1
Question 11
11.1 You have to select a new password for your "Dropbox" account on your computer. The password must consist of 3 digits and 2 letters of the alphabet in that order... show full transcript
Worked Solution & Example Answer:11.1 You have to select a new password for your "Dropbox" account on your computer - NSC Mathematics - Question 11 - 2017 - Paper 1
Step 1
11.1 The number of different passwords possible
Answer
To determine the number of possible passwords, we separate the components of the password: 3 digits and 2 vowels. Since the digit 0 is not allowed and any digit may be repeated, we have the digits 1 through 9 to consider, giving us 9 options per digit.
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Digits: There are 9 choices for each of the 3 digits, so:
-
Vowels: The vowels considered here (from the alphabet) are A, E, I, O, U (5 vowels). Since no vowels may be repeated:
- For the first vowel: 5 choices
- For the second vowel: 4 choices
Hence, the number of combinations for vowels will be:
Putting this together, the total number of passwords would be:
Step 2
11.2.1 The number of unique 12 letter arrangements that can be formed
Answer
The word 'FUNDAMENTALS' consists of 12 letters where the letter 'A' appears twice and 'N' appears twice. The formula for permutations of multiset arrangements is given by:
rac{n!}{n_1! imes n_2! imes ...}
Where:
- n is the total number of items to arrange.
- n1, n2,... are the frequencies of the repeated items.
Here,
- n = 12 (total letters)
- n1 = 2 (for A)
- n2 = 2 (for N)
Thus, the calculation becomes:
P = rac{12!}{2! imes 2!} = rac{479001600}{4} = 119750400
Step 3
11.2.2 The probability that a new arrangement will start and end with the letter N
Answer
To calculate the probability that an arrangement starts and ends with N, we can fix the N's at both ends and arrange the remaining 10 letters (which include another N). The remaining letters are: F, U, D, A, M, E, T, A, L, S.
This provides us with 10 total characters, where 'A' appears twice:
P = rac{10!}{2!} = rac{3628800}{2} = 1814400
The probability of this event occurring compared to the total arrangements calculated previously is:
P( ext{N at start and end}) = rac{1814400}{119750400} = rac{1}{66} ext{ or } 0.015
Step 4
11.3.1 Draw a tree diagram to show all possible outcomes of the coin toss
Answer
The tree diagram shows the two initial outcomes of the coin flip:
Flip
/ \
Heads Tails
For each flip, we can imagine deciding whether it is male or female flipping the coin:
M F
/ \
Heads Tails
Thus, the complete tree diagram encompasses the potential outcomes Spring:
- Male -> Heads
- Male -> Tails
- Female -> Heads
- Female -> Tails
Step 5
Step 6
11.3.3 Determine the probability that it will be a man who flips a head
Answer
To find the probability of a man flipping heads, we combine the probabilities:
- The probability of selecting a man:
P(M) = rac{1}{3} 2. The probability of flipping heads, which is:
P(H) = rac{1}{2}
Therefore, the combined probability is:
P(M ext{ and } H) = P(M) imes P(H) = rac{1}{3} imes rac{1}{2} = rac{1}{6}