What are the answers and how do you solve them?

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2 Answers
May 15, 2018

The odds are 1/10, 3/10, 6/10, in that order. It's a dependent event.

Explanation:

P(blue,blue) is equal to the odds of picking blue the first time times the odds of picking blue the second time. The first time, the odds of picking blue is 25 since there are 2 blue marbles but 5 total marbles.

If you pick a blue marble, there is 1 blue marble remaining, and 4 total marbles remaining, making the odds of picking a blue marble on the second try equal to 14

2514=220=110, so P(blue,blue)=1/10

P(red,red) involves a similar approach. The first time, the odds of picking a red marble is 35 since there are 3 red marbles, and 5 marbles total.

If you pick the red marble, there is 2 red marbles remaining, and 4 total marbles remaining, making the odds of picking a red marble on the second try equal to 24

3524=620=310

The last one you can do out step by step, following a similar method above, but there is a shortcut you can use for this case.

The odds of picking two blue marbles is 110, and the odds of picking two red marbles is 310. The odds of picking two marbles of the same color in a row is 410 then, or 25

Therefore the odds of picking two different marble colors (P(one blue, one red) is 125=35

There are two possibilities for P(one blue, one red) to happen, one of which is P(blue, red; in that order).

The odds of picking a blue marble first is 25, and the odds of picking a red marble after that is 34. Therefore the probability of the event taking place is 2534=310, meaning P(blue, red; in that order) is 310

So in conclusion

P(blue,blue)=110
P(red,red)=310
P(one blue, one red)=610
P(blue, red; in that order)=310

May 15, 2018

110,310,310,610 and dependent.

Explanation:

This is a dependent event as the second one is based after which one you took in the first one.

Knowing that, we calculate the chance of getting the first one multiplied by the chance of getting the second one from the remaining ones
2514=110 (B, B)
3524=310 (R,R)
2534=310 (B, R)
Getting one of each is also the sum of the chance of either outcome
25310+3524=610 (1B, 1R)
We can check that this is correct as the probabilites of (B, B), (R, R), (B,R) and (R, B) add up to 1