 ## Question

secondary 2 | Maths

##### John Mered

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Pls help having a hard time

Date Posted: 5 months ago
Views: 29
Eric Nicholas K
5 months ago
IFor parts a and b, I recommend drawing the possibility diagram which should have a total of 36 possible outcomes.

I will proceed without the diagram.

For the first one, if you have done the diagram, there will be 3 cases where the sum of the numbers is less than 4 (namely 1 1, 1 2 and 2 1) and 33 cases where the sum of the numbers is 4 or more. So the possibility that the numbers add up to at least 4 is 33/36.

For the second one, you will need to draw a second possibility diagram, this time for the products of the two numbers on the dice.

There are 3 cases in which the product is 24 or more, namely 4 x 6, 5 x 5 and 6 x 4. All other 33 cases result in products lower than 24.

So the possibility that the numbers multiply to less than 24 is 33/36 = 11/12.

The third one is straightforward. There are only six cases in which both numbers are the same, namely 1 1 and so on up to 6 6. The other 30 cases allow for different numbers being multiplied together. So the possibility that the two colours are different is 30/36.
Nalin Sharma
5 months ago
No, 33/36 is indeed 11/12, but for the second part there are six possible scenarios where it's more than 24, 4 and 6, 6 and 4, 5 and 5, 5 and 6, 6 and 5, 6 and 6. Hence its 30/36 which is 5/6. Same for the last part of this question.

Did not account for 5 and 5 in my working above, but that's about the only fault there. 