Mathematics, 13.07.2019 01:00 joey333

An experiment consists of drawing 10 cards from an ordinary 52-card deck. recall this deck consists cards of 13 different denominations: a, 2-10, j, q, k and four suits if the drawing is made with replacement, \ if the drawing is made without replacement, find the probability that at least 9 cards will have the same suit.

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This will be the sum of the probability that exactly 10 have the same suit plus the probability that 9 of 10 have the same suit.

With replacement first. Let's focus on a single suit, say hearts.

The probability of drawing on a heart on the first draw is 1/4. Since we're replacing, that's the probability of a heart for cards two through ten. So the probability of all hearts is $(\frac 1 4)^{10}$ .

Since there are four suits, the probability of drawing all the same suit is four times the above, $(\frac 1 4)^9$ .

The probability of drawing the first 9 hearts then a different suit for the 10th is $(\frac 1 4)^9 \cdot \frac 3 4$ .

The different suit may be any of the cards so we multiply this by 10 to get the probability of 9 of 10 hearts: $10 (\frac 1 4)^9 \cdot \frac 3 4$ .

We can have any suit, so we multiply this by a factor of 4, giving $30 (\frac 1 4)^9$ .

The probability we seek is the sum P(10 the same) + P(9 the same)

$p = (\frac 1 4)^9 + 30 (\frac 1 4)^9 = 31 (\frac 1 4)^9 = \dfrac{31}{262144}\approx 0.00012 = 0.012\%$

P(with replacement) = 0.012%

Now let's do it again without replacement.

P(Exactly 10 hearts). We have a 13/52 chance of picking a heart on the first card. Given we did, we have a 12/51 chance of picking a heart on the second card. So for all 10, we get

$\frac{13}{52} \cdot \frac{12}{51} \cdot \frac{11}{50} \cdots \frac{4}{43}$

Let's write that as

$\frac{13!}{3!} \cdot \frac{42!}{52!}$

Again the four suits gives us

$P(\textrm{All 10 the same suit}) = 4 \cdot \frac{13!}{3!} \cdot \frac{42!}{52!}$

For 9 of 10, the last fraction 4/43 becomes 39/43, then we multiply by 10 because any of the cards might not match and 4 to cover the suits.

$P(\textrm{9 of 10 the same suit}) = 4 \cdot 10 \cdot \frac{13}{52} \cdot \frac{12}{51} \cdot \frac{11}{50} \cdots \frac{5}{44} \cdot \frac{39}{43}$

$= 40 \cdot 39 \cdot \frac{13!}{4!} \cdot \frac{42!}{52!}$

So a total probability of

$p = 4 \cdot \frac{13!}{3!} \cdot \frac{42!}{52!} + 40 \cdot 39 \cdot \frac{13!}{4!} \cdot\frac{42!}{52!} = \dfrac{197}{27657385
} \approx .00071\%$

P(without replacement) = 0.00071%

A lot smaller.

I'd have to check this pretty hard to be sure it was right, which I haven't done.

Answer from: Quest

Why are u not just putting the numbers this is to much

Answer from: Quest

let x be the money she spent so that she can earn the free t-shirt.

since, she already spent $112.49

⇒ $x + 112.49$

also, as per the statement: jennifer must spend $250 or more in order to qualify for a free t-shirt at the store.

then,

we have an inequality i.,e

$x + 112.49 > 250$

subtract 112.49 from both sides we get;

$x > 137.51$

therefore, $137. 51 more money does jennifer need to spend so that she can earn the free t-shirt

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An experiment consists of drawing 10 cards from an ordinary 52-card deck. recall this deck consists...