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Mathematics 17 Online
OpenStudy (abbycross167):

Can someone please help with this math question? Please? I'll fan and give a medal...

OpenStudy (abbycross167):

OpenStudy (abbycross167):

Based on the histogram, calculate P(red, purple).

OpenStudy (mjdennis):

Okay, you will need to translate your class' shorthand. P() means "probability of" but what does it mean to have two things inside the P() operator?

OpenStudy (abbycross167):

I have to find the experimental probability

OpenStudy (abbycross167):

how do I do that with two data? would I add red and purple together? 4+2=6? then divide 6 by 16(the total times spinned)

OpenStudy (mjdennis):

That is what I don't know. Hang on

OpenStudy (mjdennis):

P(red,purple) could mean \[P(red U purple)\] or \[P(red THEN purple)\]

OpenStudy (mjdennis):

"U" means "or"

OpenStudy (mjdennis):

Let's do both if you don't know what the notation means.

OpenStudy (mjdennis):

Do you know how to read the histogram? Some outcomes happen more often than others. Looks like there are 7 green locations on the spinner, but only one orange one.

OpenStudy (mjdennis):

So first, how many red outcomes are there? How many purple outcomes are there?

OpenStudy (abbycross167):

yes sir there are 4 red and 2 purple

OpenStudy (mjdennis):

Yep yep, and how many total choices (all colors) are there?

OpenStudy (abbycross167):

16 total number of spins

OpenStudy (mjdennis):

To find P(red), we take (#red outcomes) / (total outcomes), or 4/16, or reduced to 1/4. What is P(purple) ?

OpenStudy (mjdennis):

Once we have those two, we can do the rest. Which do you want to try first, P(red OR purple), or P(red AND purple)

OpenStudy (abbycross167):

2/16 = 1/8

OpenStudy (mjdennis):

So, if we want the probability of red then purple, we can call it P(red AND purple), and in this problem we can solve by multiplying the two probabilities. P(red AND purple) = P(red) * P(purple) I am guessing you know how to multiply the two fractions?

OpenStudy (abbycross167):

yes sir

OpenStudy (mjdennis):

OK, I will be offline for a bit. If it turned out you needed the _other_ answer, ping me and I'll check in later.

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