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3 years ago

15

Please help....What is a problem that makes sense for this question? ASAP!!

1 answer:

ddd [48]3 years ago

7 0

Explanation:

250 can be used as a total, with 10 being how much something subtracted multiple times takes away, and 20 being the end result. So, we could make the problem about a lot of things, but I'll go ahead and write the first one that comes to mind.

Answer:

Arata has $250 in his bank account, and plans on buying several CDs. Each one costs $10, and by the end of the trip, there is only $20 left in his account. How many CDs did Arata buy, assuming there was no tax or other additional charges?

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What is the probability of drawing the compliment of a king or a

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Answer:

The probability of drawing the compliment of a king or a queen from a standard deck of playing cards = 0.846

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Let 'S' be the sample space associated with the drawing of a card

n (S) = 52C₁ = 52

Let E₁ be the event of the card drawn being a king

$n( E_{1} ) = 4 _{C_{1} } = 4$

Let E₂ be the event of the card drawn being a queen

$n( E_{2} ) = 4 _{C_{1} } = 4$

But E₁ and E₂ are mutually exclusive events

since E₁ U E₂ is the event of drawing a king or a queen

<u><em>step(ii):-</em></u>

The probability of drawing of a king or a queen from a standard deck of playing cards

P( E₁ U E₂ ) = P(E₁) +P(E₂)

$= \frac{4}{52} + \frac{4}{52}$

P( E₁ U E₂ ) = $\frac{8}{52}$

<u><em>step(iii):-</em></u>

The probability of drawing the compliment of a king or a queen from a standard deck of playing cards

$P(E_{1}UE_{2}) ^{-} = 1- P(E_{1} U E_{2} )$

$P(E_{1}UE_{2}) ^{-} = 1- \frac{8}{52}$

$P(E_{1}UE_{2}) ^{-} = \frac{52-8}{52} = \frac{44}{52} = 0.846$

<u><em>Conclusion</em></u>:-

The probability of drawing the compliment of a king or a queen from a standard deck of playing cards = 0.846

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