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Reptile [31]
2 years ago
10

Urgent. There are 12 coloured counters in a bag. The counters are black, white or grey A counter is chosen at random. The probab

ility that the counter is not black is 3/4. The probability that the counter is not white is 2/3. Work out the probability that the counter is grey.​
Mathematics
1 answer:
Nastasia [14]2 years ago
7 0

Answer:

1/12

Step-by-step explanation:

<u>Needed information</u>

\sf Probability\:of\:an\:event\:occurring = \dfrac{Number\:of\:ways\:it\:can\:occur}{Total\:number\:of\:possible\:outcomes}

The sum of the probabilities of all outcomes must equal 1

<u>Solution</u>

We are told that the probability that the counter is <em>not</em> black is 3/4.  

As the sum of the probabilities of all outcomes <u>must equal 1</u>, we can work out the probability that the counter <em>is </em>black by subtracting 3/4 from 1:

\sf P(counter\:not\:black)=\dfrac34

\implies \sf P(counter\:black)=1-\dfrac34=\dfrac14

We are told that the probability that the counter is <em>not </em>white is 2/3.  

As the sum of the probabilities of all outcomes <u>must equal 1</u>, we can work out the probability that the counter <em>is </em>white by subtracting 2/3 from 1:

\sf P(counter\:not\:white)=\dfrac23

\implies \sf P(counter\:white)=1-\dfrac23=\dfrac13

We are told that there are black, white and grey counters in the bag.  We also know that the sum of the probabilities of all outcomes must equal 1.  Therefore, we can work out the probability the counter is grey by subtracting the probability the counter is black and the probability the counter is white from 1:

\begin{aligned}\sf P(counter\:grey) & = \sf1-P(counter\:black)-P(counter\:white)\\\\ & =\sf 1-\dfrac14-\dfrac13\\\\ & = \sf \dfrac{12}{12}-\dfrac{3}{12}-\dfrac{4}{12}\\\\ & = \sf \dfrac{12-3-4}{12}\\\\ & = \sf \dfrac{5}{12}\end{aligned}

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