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pav-90 [236]
1 year ago
15

coin $a$ is flipped three times and coin $b$ is flipped four times. what is the probability that the number of heads obtained fr

om flipping the two fair coins is the same?
Mathematics
1 answer:
ivolga24 [154]1 year ago
4 0

The probability that the number of heads obtained from flipping the two fair coins is the same is 35/128.

Probability:

Probability means the fraction of favorable outcome and the total number of outcomes.

So it can be written as,

Probability = Favorable outcomes / Total outcomes

Given,

The coin a is flipped three times and coin b is flipped four times.

Here we need to find the probability that the number of heads obtained from flipping the two fair coins is the same.

We know that,

There are 4 ways that the same number of heads will be obtained;

0, 1, 2, or 3 heads.

The probability of both getting 0 heads is

$\left(\frac12\right)^3{3\choose0}\left(\frac12\right)^4{4\choose0}=\frac1{128}$

Probability of getting 1 head,

$\left(\frac12\right)^3{3\choose1}\left(\frac12\right)^4{4\choose1}=\frac{12}{128}$

Probability of getting 2 heads is,

$\left(\frac12\right)^3{3\choose2}\left(\frac12\right)^4{4\choose2}=\frac{18}{128}$

And the probability of getting 3 heads is,

$\left(\frac12\right)^3{3\choose3}\left(\frac12\right)^4{4\choose3}=\frac{4}{128}$

Therefore, the probability that the number of heads obtained from flipping the two fair coins is the same is,

=> (1/128) + (12/128) + (18/128) + (4/128)

=> 35/128.

To know more about probability here

brainly.com/question/14210034

#SPJ4

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Dovator [93]

Answer:

Answers are listed hoping that they make sense XD

Step-by-step explanation:

1. -0.5

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I just substituted the value given from x and solved for y. I hope that is what the question is asking for.

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4 0
3 years ago
For a set of five whole numbers, the mean is 4, the mode is 1, and the median is 5. What are the five numbers?
zmey [24]
SO we need 5 numbers that equal to 20, with 5 being the middle number, and ones being the most used number so

1 1 5 x y

According to this, x and y must equal 13 and both be bigger then 5. Because we can rule out every other number, x and y must be 6 and 7.

Your 5 numbers are 1, 1, 5, 6, 7.
5 0
3 years ago
BCALLE can u help me please
Ksivusya [100]
17. RQ is the same as PS.
PS = -1 + 4x
RQ = 3x + 3
-1 + 4x = 3x + 3
4x = 3x + 4
x = 4
Now plug that into RQ.
3(4) + 3 = RQ
15 = RQ

18. Angles G and E are equal to each other.
G = 5x - 9
E = 3x + 11
5x - 9 = 3x + 11
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2x = 20
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Plug that x into G.
5(10) - 9
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19. TE and EV are equal to each other.
TE = 4 + 2x
EV = 4x - 4
4 + 2x = 4x - 4
2x = 4x - 8
-2x = -8
x = 4
Plug that into TE.
4 + 2(4)
12 = TE

20. DB and BF are equal.
DB = 5x - 1
BF = 5 + 3x
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4 0
3 years ago
Tim Worker decided to purchase a new DVD player on an installment loan. The DVD player was $295.00. Tim agreed to pay $31.00 per
Kryger [21]
Hi there

Total paid
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finance charge
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Hope it helps
5 0
3 years ago
A box contains 5 red balls, 6 white balls and 9 black balls. Two balls are drawn at
valina [46]

Answer:

P(Same)=\frac{61}{190}

Step-by-step explanation:

Given

Red = 5

White = 6

Black = 9

Required

The probability of selecting 2 same colors when the first is not replaced

The total number of ball is:

Total = 5 + 6 + 9

Total = 20

This is calculated as:

P(Same)=P(Red\ and\ Red) + P(White\ and\ White) + P(Black\ and\ Black)

So, we have:

P(Same)=\frac{n(Red)}{Total} * \frac{n(Red) - 1}{Total - 1} + \frac{n(White)}{Total} * \frac{n(White) - 1}{Total - 1}  + \frac{n(Black)}{Total} * \frac{n(Black) - 1}{Total - 1}

<em>Note that: 1 is subtracted because it is a probability without replacement</em>

P(Same)=\frac{5}{20} * \frac{5 - 1}{20- 1} + \frac{6}{20} * \frac{6 - 1}{20- 1}  + \frac{9}{20} * \frac{9- 1}{20- 1}

P(Same)=\frac{5}{20} * \frac{4}{19} + \frac{6}{20} * \frac{5}{19}  + \frac{9}{20} * \frac{8}{19}

P(Same)=\frac{20}{380} + \frac{30}{380}  + \frac{72}{380}

P(Same)=\frac{20+30+72}{380}

P(Same)=\frac{122}{380}

P(Same)=\frac{61}{190}

4 0
3 years ago
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