A bag contains n metal coins, n ≥ 3, that are made from either silver or bronze - HSC - SSCE Mathematics Extension 1 - Question 9 - 2024 - Paper 1
Question 9
A bag contains n metal coins, n ≥ 3, that are made from either silver or bronze.
There are k silver coins in the bag and the rest are bronze.
Two coins are to be d... show full transcript
Worked Solution & Example Answer:A bag contains n metal coins, n ≥ 3, that are made from either silver or bronze - HSC - SSCE Mathematics Extension 1 - Question 9 - 2024 - Paper 1
Step 1
Calculate the Total Number of Coins
96%
114 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
The total number of coins is given by n=k+(n−k), where k is the number of silver coins. Therefore, there are n−k bronze coins.
Step 2
Calculate Probability for Silver Coins
99%
104 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
The probability of drawing two silver coins can be computed as follows:
The probability of drawing the first silver coin: ( P(S_1) = \frac{k}{n} )
After drawing one silver coin, the probability of drawing another silver coin:
( P(S_2 | S_1) = \frac{k - 1}{n - 1} )
Thus, the total probability of drawing two silver coins is:
[ P(S) = \frac{k}{n} \times \frac{k - 1}{n - 1} = \frac{k(k - 1)}{n(n - 1)} ]
Step 3
Calculate Probability for Bronze Coins
96%
101 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
Similarly, the probability of drawing two bronze coins is:
The probability of drawing the first bronze coin:
( P(B_1) = \frac{n - k}{n} )
After drawing one bronze coin, the probability of drawing another bronze coin:
( P(B_2 | B_1) = \frac{n - k - 1}{n - 1} )
Thus, the total probability of drawing two bronze coins is:
[ P(B) = \frac{n - k}{n} \times \frac{n - k - 1}{n - 1} = \frac{(n - k)(n - k - 1)}{n(n - 1)} ]
Step 4
Total Probability of Drawing Two Coins of Same Metal
98%
120 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
Therefore, the total probability of drawing two coins of the same metal (either silver or bronze) is:
[ P(S) + P(B) = \frac{k(k - 1)}{n(n - 1)} + \frac{(n - k)(n - k - 1)}{n(n - 1)} ]
To simplify:
[ P(S + B) = \frac{k(k - 1) + (n - k)(n - k - 1)}{n(n - 1)} ]