A box contains 6 red balls, 4 white balls, and 5 green balls. If three balls are drawn in succession without being replaced, what is the probability that they are drawn in the order red, white, green?

A
x
8/225
B
x
4/91
C
x
12/445
D
x
1/225

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A card is pulled from a well shuffled deck of standard playing cards and is not put back in the deck. A second card is drawn from the deck. What is the probability that a spade was drawn both times?

A
x
Independent; 1/16
B
x
Dependent; 1/16
C
x
Independent; 1/17
D
x
Dependent; 1/17

1st question:

prob = (6/15)(4/14)(5/13) = 4/91

2nd problem:

prob = (13/52)(12/51) = 1/17

To find the probability of events happening in a specific order, you can multiply the probabilities of each event occurring.

For the first question:
1. Find the probability of drawing a red ball first from the box. There are a total of 15 balls in the box, and 6 of them are red. So, the probability of drawing a red ball first is 6/15.
2. After drawing a red ball, there are 14 balls left in the box, including 4 white balls. Therefore, the probability of drawing a white ball second is 4/14.
3. Now, there are 13 balls left in the box, including 5 green balls. So, the probability of drawing a green ball third is 5/13.
4. Multiply the probabilities of each event occurring: (6/15) * (4/14) * (5/13) = 8/225.
Therefore, the correct answer is A, 8/225.

For the second question:
1. Find the probability of drawing a spade for the first card. There are 13 spades in a deck of 52 playing cards, so the probability of drawing a spade first is 13/52.
2. After drawing the first card, there are now 51 cards left in the deck, including 12 spades. Therefore, the probability of drawing a spade for the second card is 12/51.
3. Multiply the probabilities of each event occurring: (13/52) * (12/51) = 1/17.
Therefore, the correct answer is D, Dependent; 1/17.