A spinner and 4 cards are shown below: A spinner with 5 equal sectors is shown in the figure. The colors Green, Blue, Purple, Yellow, and Red are marked on it. The arrow points at the purple color. Four cards are shown on the right side. The colors Red, Yellow, Blue, and Pink are marked on them. Jane spins the spinner and selects a card without looking. What is the probability that the spinner stops at purple and a pink card is selected? 1 over 20 1 over 5 1 over 4 9 over 20
here is the picture: http://learn.flvs.net/webdav/assessment_images/educator_mjmath2_v14/mjmath2_practice_m7_g1_p.jpg
@igreen
@dr0zier99
HI!!
cookie error (mmm cookies)
hello will you please help me?
love to but link does not work post a screenshot
oooh bet it is from FLVS right?
ok one second
yes it is from FLVS
I think I might know the answer but honestly this is just multiplication and addition
i know the answer
help
ooh ooh me too pick me!
tell me pls
just kidding
tell me @dr0zier99
probability of the first one being purple is \(\frac{1}{5}\) since there are 5 evenly placed sections on the spinner
yes go on
probability a pink card is \(\frac{1}{4}\) since there are 4 cards and one is pink
ok
to find the probability you get both , multiply
is it 1/20?
\[\frac{1}{5}\times \frac{1}{4}\]
yes
:) @@
these 2 events are independent of each other so the probabilities are multiplied
we are assuming that the outcomes of each selection are "independent" so you can just multiply the probabilities together
help on another question?
sure !!
imma close this one and open another ok?
you have lots of attention for this question, bet you get a lot of answers
get ready
all set!
The spinner has 5 possible outcomes: Green, Blue, Purple, Yellow, and Red The cards have 4 possible outcomes: Red, Yellow, Blue, and Pink p(spinner purple) = 1/5 p(pink card) = 1/4 p(spinner purple and pink card) = p(spinner purple) * p(pink card) = 1/5 * 1/4 = 1/20
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