I'm getting a bit dumb on the probabilities.
Tasks like:
On Monday, there is a 10% chance of rain. On Tuesday, there is a 10% chance of rain. On Wednesday, there is a 10% chance of rain. What is the probability that it will rain on one of these three days?
I'm used to solving by the "backwards method". First, we calculate the probability that it will NOT rain on one of these days 0.9*0.9*0.9=0.729, and then subtract from one this number 1-0.729=0.271. So the answer is 27.1%.
Is there a way to directly calculate the required value?
0,1+0,9*0,1+0,9*0,9*0,1=0,271
0,1+0,9*0,1+0,9*0,9*0,1=0,271
Sens. I'll look into it.
Tasks like:
On Monday, there is a 10% chance of rain. On Tuesday, there is a 10% chance of rain. On Wednesday, there is a 10% chance of rain. What is the probability that it will rain on one of these three days?
50%. Either it goes or it doesn't.
:)
50%. Either it goes or it doesn't.
:)
Even more divided. What a challenge. :)
50%. Either it goes or it doesn't.
:)
That's right.
- What are the chances of meeting a dinosaur on the street right now?
- 50%. Meeting/not meeting.
30% according to the probability addition rule (or or).
If one of the above three days goes and two days are without precipitation, then: 3 * 0.1 * 0.9 * 0.9 = 0,243
And that's the way it is in all the threads :) without any hinting and without any understanding.
It will be a pain to sort things out, because the topicwriter, and not the first, cannot clearly set out the terms of the problem. It's useless and time-consuming to get anything out of them. That's why in my reply I give one of the possible variants of the problem's conditions.
And there is generally nothing to appeal to, because everything is calculated by the banal Bernoulli formula: the probability of one success in three trials.
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Tasks like:
On Monday, there is a 10% chance of rain. On Tuesday, there is a 10% chance of rain. On Wednesday, there is a 10% chance of rain. What is the probability that it will rain on one of these three days?
I'm used to solving by the "backwards method". First, we calculate the probability that it will NOT rain on one of these days 0.9*0.9*0.9=0.729, and then subtract from one this number 1-0.729=0.271. So the answer is 27.1%.
Is there a way to directly calculate the required value?