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Readme [11.4K]
3 years ago
9

A deck of cards contains RED cards numbered 1,2,3, BLUE cards numbered 1,2,3,4,5,6, and GREEN cards numbered 1,2. If a single ca

rd is picked at random, what is the probability that the card is GREEN AND has an ODD number? Provide the final answer as a fraction.
Mathematics
2 answers:
vfiekz [6]3 years ago
8 0

Answer:

1/52

Step-by-step explanation:

A deck of cards contains 52 cards

Red cards are numbered 1,2,3

Blue cards are numbered 1,2,3,4,5,6

Green cards are numbered 1,2

Number of Red card = 3

Number of blue card = 6

Number of Green card = 2

Let Pr(R) = Probability of picking a Red card

Let Pr(B) = probability of picking a blue card

Let Pr(G) = probability of picking a green card

Let Pr(RE) = Probability of picking a red even card

Let Pr(RO) = probability of picking a red odd card

Let Pr(BE) =probability of picking a blue even card

Let Pr(BO) = probability of picking a blue odd card

Let Pr(GE) = probability of picking a green even card

Let Pr(GO) = probability of picking a green odd card

Pr(R) = 3/52

Since we have 3 red cards,

Pr(RE) = 1/3

Pr(RO) =2/3

Pr(B) = 6/52

= 3/26

Since we have 6 blue cards ,

Pr(BE) = 3/6

= 1/2

Pr(BO) = 3/6

= 1/2

Pr (G) = 2/52

= 1/26

Since we have 2 green cards,

Pr(GE) = 1/2

Pr(GO) = 1/2

The probability of picking a Green card and an odd green card is

Pr(G) n pr(GO)

1/26 * 1/2

= 1/52

irakobra [83]3 years ago
3 0

Answer:

8/15 because its 8 odd numbers out of 15 numbers all together

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Answer:

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Step-by-step explanation:

We are given the following information in the question:

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P( X > x) = P( z > \displaystyle\frac{x - 5}{1})=0.95  

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