Hi guys, I always seem to get probability questions wrong so would someone be able to give any tips on how to understand the wording of questions (like the one below), or if there's a cheat way to always get the right answer? haha thank you!!!
oh and also the second probability question is something i never learnt because I was away in year 10 when it was taught and didn't do prelim general, so if you could also explain how to do these questions that'd be much appreciated!! 
THANK YOU!!
Hey! Probability questions are always really tough, so to be honest my only advice is to do loads of practice. I did 4U maths, but still found this sort of probability really hard, particularly just understanding what the hell the question was actually asking. By doing loads of questions, and looking at the answers when you get stuck, you start to get a feel for when you should be using certain techniques.
Now, let's look at the first question. Clearly, the unknown value will be 12 (the total) - 4 (predicted success/actual failure) = 8 (predicted failure/actual failure).
Now, if a business is chosen at random, what is the probability that it succeeded? This is completely detached from the predictions, so we just look at the total business, and their actual failures/success. There are 50 total businesses, 38 of them succeeded, and 12 of them failed. So, the probabilities will be
=\frac{38}{50})
=\frac{12}{50})
Okay, now the last part is the most difficult. If she predicted success, what is the likely hood that the business actually succeeded?
Well, if she predicted success, 28 of them succeeded and 4 of them failed. So, the probability is just
=\frac{28}{28+4}=\frac{28}{32})
=\frac{4}{32})
As it will just be the number of successes, over the total number! That wasn't actually so bad, but understanding the table was tricky.
Now, let's look at the second question. Because the letter 'S' MUST be at the start, we ignore that fact (ie. there's a 100% probability that it starts with S, so we don't really care about the letter anymore). The rest of the digits (four) can be any of 10 numbers.
We think about this in steps. Ie. How many ways could we choose the first number? How many ways could we choose the second number? How many ways could we choose the third number? How many ways could we choose the fourth number? Then, we simply multiply them together to get the
total ways.
Each time we choose a number, there will be 10 numbers to choose from. So, each of the steps will have 10 choices, and the answer will be D (10*10*10*10). Easy!
The question could have been a little more complicated, though. Let's say you couldn't repeat the same number twice (ie. If I chose a 3 the first time, I couldn't choose a 3 for any other number).
First, we would have 10 choices. Then, we would have 9 choices (as one cannot be reused, which is whatever we chose the first time). Then, we have 8 choices. Then, we have 7 choices. So, our answer would be 10*9*8*7. This is just an example of how these questions commonly work.
Hope all of this makes sense!
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