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kykrilka [37]
2 years ago
14

You randomly select one card from a 52 card deck find the probability of selecting a ace or a two

Mathematics
1 answer:
balu736 [363]2 years ago
3 0
The odds of picking an ace or a 2 are 8 out of 52 or 8/52 or 2/13.
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"If two cards are drawn at random without replacement from a standard deck, find the probability that the second card is a face
PolarNik [594]

Answer:

C. P(F|Q) = \dfrac{11}{51}

Step-by-step explanation:

it is to be noted that the question is only asking for the probability of the 2nd card given that the first card was queen (P(F|Q)), and not asking for the probability of 1st card to be queen and 2nd card to be faced cardP(Q\,\text{and}\,F)

we can represent it in an expression:

P(Q\,\text{and}\,F) = P(Q)P(F|Q)

here P(Q) is the first event: Queen

and P(F|Q) is the second event: Faced card, given that the Queen is taken

--------------------------------

we only need to know what is P(F|Q), and that can be found directly found:

let's start with P(Q), what is the probability that the first card is a Queen? Well, there are 4 queens in a standard deck of 52 cards, so the probability should be:

P(Q) = \dfrac{4}{52}

now we have taken our queen, but we haven't put it back in the deck. so the amount of cards in the deck now are 51.

let's calculate P(F|Q),now that one queen is taken out, what is the probability of the next card to be a faced card? Well, in a standard deck there are 12 faced cards, but in our case one queen is already taken out, so there are 11 faced cards in our deck!

P(F|Q) = \dfrac{11}{51}

and this our answer!

8 0
3 years ago
(32 minus 20) divided by 4
aalyn [17]
The answer is 3. I hope this helps!
8 0
4 years ago
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What are all the possible cross sections of a cone
natita [175]

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big

Step-by-step explanation:

2

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3 years ago
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Answer:

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3 0
3 years ago
Company 1: 3.4 defects per 980,000 occurrences Company 2: 4 defects per 1,115,783 occurrences Company 3: 18 defects per 5,999,52
Gelneren [198K]

Given Information:  

Company 1: defects = 3.4

Company 1: units = 980,000

Company 2: defects = 4

Company 2: units = 1,115,783

Company 3: defects = 18

Company 3: units = 5,999,521

Required Information:  

six-sigma level = ?

Answer:

Company 1: Possible

Company 3: Impossible

Company 3: Possible

Step-by-step explanation:

The six sigma levels are from 1 to 6 where 1 represents lowest quality level and 6 represents highest quality level

To find out sigma level first we have to find out DPU and DMPU

Where DPU is no. of defects divided by no. of units

And DMPU is DPU multiplied by 1 million

Company 1:

DPU = 3.4/980,000 = 0.0000030612

DMPU = 0.0000030612*1,000,000 = 3.06

Company 2:

DPU = 4/1,115,783 = 0.000003584

DMPU = 0.000003584*1,000,000 = 3.58

Company 3:

DPU = 18/5,999,521 = 0.00000300

DMPU = 0.00000300*1,000,000 = 3.0

According to six sigma tables, for six-sigma level, DMPU must be equal or less than 3.4. As you can see only company 1 and 3 has DMPU less than 3.4 therefore, it is possible for company 1 and company 3 to meet six-sigma level but impossible for company 2.

3 0
3 years ago
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