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Imagine that you have a bag with 2 red blocks, 3 blue blocks, 4 green blocks, and 1 yellow block inside.
You plan to make one draw from the bag.

A Find the probability of getting a green block or a blue block. p(blue or green)=??? (Do not Simplify any Fraction)

you plan on making two draws from the bag (replacing the first)
B Find the probability of getting a green block then a blue block. (order matters) p(green then blue) = ???

C Which is more likely to happen? A or B? Write the letter down.


Sagot :

Answer:

it’s B

Step-by-step explanation:

Answer:

Information given

3 blue

2 red

4 green

1 yellow

Total= 10

For (1)  for the probability of a green OR blue blocks you jsut gotta add up green and blue. 4+3=7 And 10 blocks in total so 7/10 chance

For (b) you find the probability of a green which is 4/10 and then a blue which is 3/10. So assumingg you draw twice you have 4/10*3/10 which is 12/100 chance of getting that order.

Well of course b is most likely to happen because its the most Chances.

7/10 > 12/100

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