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KonstantinChe [14]
3 years ago
5

Gloria had two sets of alphabet cards (A–Z). She mixed the two sets together to form a single stack of cards. Then Gloria drew t

hree cards from the stack without replacing them. What is the probability that she drew either a card with an L or a card with an R all three times?
Mathematics
1 answer:
grin007 [14]3 years ago
3 0
She has 52 card all together so she has a probability of drawing on the first draw 4 L's or R's because their are two sets of alphabet cards.  So that is 4/52 or 1/13 to the lowest term.  Second turn she only has 51 cards to draw from and still has 4 L's and R's so that would be 4/51 and on the third try she has only 50 cards left so that would be 4/50 or 2/25 to the lowest term.  Now multiply all three factions 1/13 x 4/51 x 2/25 = 8/16575 meaning out of the three draws she has a probability of getting a L or R, 8 out of 16575 each draw.
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madam [21]

Answer:

The probability of selecting a red  marble, not replacing it, and then selecting a green marble from the bag is 24/95

Step-by-step explanation:

Number of red marbles = 12

Number of green marbles = 8

Total number of marbles = 12+8 = 20

Probability of selecting red marble =\frac{12}{20}

Since it is the case of no replacement

Remaining marbles = 20-1 = 19

Number of red marbles = 12-1=11

Number of green marbles = 8

Probability of selecting green marble =\frac{8}{19}

So, the probability of selecting a red  marble, not replacing it, and then selecting a green marble from the bag =\frac{12}{20} \times \frac{8}{19}=\frac{24}{95}

Hence the probability of selecting a red  marble, not replacing it, and then selecting a green marble from the bag is 24/95

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Likurg_2 [28]

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Step-by-step explanation:

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2/3 x 10/10 = 20/30

3/10 x 3/3  = 9/30

20/30 + 9/30 = 29/30

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