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30 students have entered a poetry competition.
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One student will win the Golden Prize and another student will win the Silver Prize.
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Part a) How many different possible combinations of Golden Prize and Silver Prize winners are there in total?
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There is a second part to this question, which we will look at later.
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In order to solve this problem, we’re looking for the different combinations of Golden Prize and Silver Prize winners.
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As there’re 30 students in total, the probability or chance of winning the Golden Prize is one out of 30.
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This means that there are 30 possible Golden Prize winners.
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Once the Golden Prize winner has been awarded, there is a one out of 29 chance for the remaining students of winning the Silver Prize.
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This means that for each of the Golden Prize winners, there are 29 possible Silver Prize winners.
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We can show this by firstly numbering each of the students from one to 30.
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If student number one wins the Golden Prize, then any of the other 29 students could win the Silver Prize.
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It could be student number two, number three, number four, and on so on all the way to number 30.
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This gives us 29 combinations.
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If student number two won the Golden Prize, then student number one, student number three, four, five, and so on could win the Silver Prize.
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Once again, this gives us 29 combinations.
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We could repeat this with student number three, four, five, and so on winning the Golden Prize up to student number 30.
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This means that for each of the 30 possible Golden Prize winners, there’re 29 possible Silver Prize winners.
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We need to multiply 30 by 29.
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This is equal to 870 as three multiplied by 29 is 87.
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There were, therefore, 870 different possible combinations of Golden Prize and Silver Prize winners.
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The second part of the question says the following.
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For a game, two opaque bags containing coloured balls are used.
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In the red bag, the 14 balls are red and numbered from one to 14.
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In the yellow bag, the 11 balls are yellow and numbered from one to 11.
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The game consists of picking a ball at random from one of the bags and then another ball from either the same bag or the other bag.
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Each round of the game results in a list of two colour-number pairs.
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Part b) How many different lists can be created with this game?
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The key start point here is that we are selecting two balls.
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We could select two red balls, a red ball followed by a yellow ball, a yellow ball followed by a red ball, or two yellow balls.
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In total, we’ve 14 red balls and 11 yellow balls.
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The probability of picking a specific red ball would be one out of 14.
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And picking a different specific red ball on our second pick would be one out of 13.
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This means to calculate the number of different combinations of two red balls.
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We need to multiply 14 by 13.
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This is equal to 182.
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Next, we will look at a red ball followed by yellow ball.
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Once again, the probability of picking a specific red ball is one out of 14 and the probability of picking a specific yellow ball is one out of 11.
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The total number of combinations is then calculated by multiplying 14 by 11.
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This is equal to 154.
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Our third option was a yellow followed by a red.
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This is calculated in the same way: one out of 11 and one out of 14.
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11 multiplied by 14 is also 154.
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Finally, we need to consider two yellow balls.
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Picking two balls from this bag would be one out of 11 and one out of 10.
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This means that the number of different combinations of two yellow balls is 110 as 11 multiplied by 10 is 110.
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Adding these four numbers gives us 600.
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Therefore, we can say that there are 600 different lists that can be created with this game.
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In other words, there are 600 different combinations of balls that can be selected.