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There are only blue counters, yellow counters, green counters and red counters in a bag - Edexcel - GCSE Maths - Question 10 - 2017 - Paper 3

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There are only blue counters, yellow counters, green counters and red counters in a bag. A counter is taken at random from the bag. The table shows the probabilitie... show full transcript

Worked Solution & Example Answer:There are only blue counters, yellow counters, green counters and red counters in a bag - Edexcel - GCSE Maths - Question 10 - 2017 - Paper 3

Step 1

Work out the probability of getting a red counter.

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Answer

To find the probability of getting a red counter, we first recognize that the sum of the probabilities of all possible outcomes must equal 1. Therefore, we can set up the equation:

P(blue)+P(yellow)+P(green)+P(red)=1P(blue) + P(yellow) + P(green) + P(red) = 1

Substituting the known probabilities:

0.2+0.35+0.4+P(red)=10.2 + 0.35 + 0.4 + P(red) = 1

Calculating the sum of the probabilities of blue, yellow, and green:

0.2+0.35+0.4=0.950.2 + 0.35 + 0.4 = 0.95

Now substituting back into the equation:

0.95+P(red)=10.95 + P(red) = 1 P(red)=10.95=0.05P(red) = 1 - 0.95 = 0.05

Thus, the probability of getting a red counter is 0.05.

Step 2

What is the least possible number of counters in the bag?

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Answer

To find the least possible number of counters in the bag, we need to consider the probabilities calculated previously. The probability of getting a red counter is 0.05, which can be expressed in terms of the total number of counters, denoted as N.

Therefore, we have:

P(red)=number of red countersN=0.05P(red) = \frac{number\ of\ red\ counters}{N} = 0.05

This implies that the number of red counters must be at least:

number of red counters=0.05Nnumber\ of\ red\ counters = 0.05N

Since the number of counters must be a whole number, we can rearrange the equation. For the smallest whole number, N must be a multiple of 20 (since 0.05 of 20 is 1).

Thus, the least possible number of counters in the bag that satisfies this condition is 20.

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