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alexgriva [62]
3 years ago
9

Steve has 8 biscuits in a tin. There are 5 digestive and 3 chocolate biscuits. Steve takes two biscuits at random from the tin.

Work out the probability that he chooses two different types of biscuits.
Mathematics
2 answers:
Talja [164]3 years ago
8 0

Answer:

15/28

Step-by-step explanation:

Fudgin [204]3 years ago
5 0

Answer:

5/7

Step-by-step explanation:

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<span>F for Frank, A or Alice. F(initial)=1.95 inches A(initial)=1.50 inches Frank's equation at .25 inches per year and t representing year variable. F=1.95+.25t Alice's equation at .40 inches per year and t representing year variable. A=1.5+.40t To figure out how old they will be when their beaks are the same lengths set the equations equal to eachother as the equations are length. 1.95+.25t=1.5+.40t .45=.15t t=3 years</span>
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1) Our marbles will be blue, red, and green. You need two fractions that can be multiplied together to make 1/6. There are two sets of numbers that can be multiplied to make 6: 1 and 6, and 2 and 3. If you give the marbles a 1/1 chance of being picked, then there's no way that a 1/6 chance can be present So we need to use a 1/3 and a 1/2 chance. 2 isn't a factor of 6, but 3 is. So we need the 1/3 chance to become apparent first. Therefore, 3 of the marbles will need to be one colour, to make a 1/3 chance of picking them out of the 9. So let's say 3 of the marbles are green. So now you have 8 marbles left, and you need a 1/2 chance of picking another colour. 8/2 = 4, so 4 of the marbles must be another colour, to make a 1/2 chance of picking them. So let's say 4 of the marbles are blue. We know 3 are green and 4 are blue, 3 + 4 is 7, so the last 2 must be red.
The problem could look like this:

A bag contains 4 blue marbles, 2 red marbles, and 3 green marbles. What are the chances she will pick 1 blue and 1 green marble?

You should note that picking the blue first, then the green, will make no difference to the overall probability, it's still 1/6. Don't worry, I checked

2) a - 2%  as a probability is 2/100, or 1/50. The chance of two pudding cups, as the two aren't related, both being defective in the same packet are therefore 1/50 * 1/50, or 1/2500.  

b - 1,000,000/2500 = 400
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