?Ten slips of paper are each marked with a different number (from 0 to 9) and placed in a basket. A slip is pulled out, its number is recored (in the order it is drawn), and the slip is replaced. This is done 4 times.

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?Ten slips of paper are each marked with a different number (from 0 to 9) and placed in a basket. A slip is pulled out, its number is recored (in the order it is drawn), and the slip is replaced. This is done 4 times.

How to find the probabilities that the follow (numbers) are formed?

a) 1234
b) a number which starts with 5
c) a number with no repetition of digits.
d) a number that does not contain 2,3, or 4

2 Answers

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Answer 1 · 2017-08-07T19:38:16

As explained below.

Explanation:

There are $10$ $10$ different digits. The probability of drawing any number is $\frac{1}{10}$ $1/10$

a) To form the number $1234$ $1234$ , no choice is allowed.
The numbers must be drawn in this order.

$P (1) \times P (2) \times P (3) \times P (4) = \frac{1}{10} \times \frac{1}{10} \times \frac{1}{10} \times \frac{1}{10}$ $P(1) xx P(2) xx P(3) xx P(4) = 1/10xx 1/10xx1/10xx1/10$

$= \frac{1}{10, 000}$ $= 1/(10,000)$

The total number of ways of drawing $4$ $4$ cards is:

$10 \times 10 \times 10 \times 10 = 10, 000$ $10xx10xx10xx10 = 10,000$

b) To form a number which start with $5$ $5$ , there is only $1$ $1$ digit allowed for the first card.

However, the numbers which follow can be any of the $10$ $10$ digits.

$P (5 N N N) = \frac{1000}{10000} = \frac{1}{10}$ $P(5N N N) = 1000/10000 =1/10$
This can also be calculated as $\frac{1}{10} \times \frac{10}{10} \times \frac{10}{10} \times \frac{10}{10} = \frac{1}{10}$ $1/10 xx10/10xx10/10xx10/10 =1/10$

c) If no repetition is allowed, then the number of choices drops with each successive draw.
There are $10$ $10$ choices for the first, then, $9$ $9$ , then $8$ $8$ , then $7$ $7$ .

Number of ways $= 10 \times 9 \times 8 \times 7 = 5040$ $= 10xx9xx8xx7 = 5040$
(This allows for the first digit to be $0$ $0$ )

$\frac{10}{10} \times \frac{9}{10} \times \frac{8}{10} \times \frac{7}{10} = \frac{5040}{10000}$ $10/10 xx9/10xx8/10xx7/10 = 5040/10000$

If the first digit may not be $0$ $0$ ,
then the number of ways $= 9 \times 9 \times 8 \times 7 = 4536$ $= 9xx9xx8xx7 =4536$

$\frac{9}{10} \times \frac{9}{10} \times \frac{8}{10} \times \frac{7}{10} = \frac{4536}{10000}$ $9/10xx9/10xx8/10xx7/10 = 4536/10000$

$P (no digit repeating) = \frac{5040}{10000} or \frac{4536}{10000}$ $P("no digit repeating") = 5040/10000 or 4536/10000$

d) If a number may not contain $2, 3, 4$ $2,3,4$ then there are only $7$ $7$ choices.

Number of ways is $7 \times 7 \times 7 \times 7 = 2401$ $7xx7xx7xx7 = 2401$

If the first digit may be $0$ $0$

$\frac{7}{10} \times \frac{7}{10} \times \frac{7}{10} \times \frac{7}{10} = \frac{2401}{10000}$ $7/10xx7/10xx7/10xx7/10 = 2401/10000$

$P (not 2,3,4) = \frac{2401}{10000}$ $P("not 2,3,4") = 2401/10000$

If the first digit may not be $0$ $0$ , then
the number of ways $6 \times 7 \times 7 \times 7 = 2058$ $6xx7xx7xx7 = 2058$

$P (not 2,3,4) = \frac{2058}{10000}$ $P("not 2,3,4") = 2058/10000$

Answer 2 · 2017-08-07T20:04:38

$a . \frac{1}{10000} = 0.0001$ $a. 1/10000=0.0001$

$b . \frac{1}{10} = 0.1$ $b. 1/10=0.1$

$c . \frac{504}{1000} = \frac{63}{125} = 0.504$ $c. 504/1000=63/125=0.504$

$d . \frac{2401}{10000} = 0.2401$ $d. 2401/10000=0.2401$

Explanation:

a. $P (Any card) = \frac{1}{10}$ $P("Any card")=1/10$

Therefore, $P (1) = \frac{1}{10}$ $P(1)=1/10$

$P (12) = (\frac{1}{10}) \cdot (\frac{1}{10}) = {(\frac{1}{10})}^{2} = \frac{1}{100}$ $P(12)=(1/10)*(1/10)=(1/10)^2=1/100$

$P (123) = (\frac{1}{10}) \cdot (\frac{1}{10}) \cdot (\frac{1}{10}) = {(\frac{1}{10})}^{3} = \frac{1}{1000}$ $P(123)=(1/10)*(1/10)*(1/10)=(1/10)^3=1/1000$

$P (1234) = (\frac{1}{10}) \cdot (\frac{1}{10}) \cdot (\frac{1}{10}) \cdot (\frac{1}{10}) = {(\frac{1}{10})}^{4} = \frac{1}{10000}$ $P(1234)=(1/10)*(1/10)*(1/10)*(1/10)=(1/10)^4=1/10000$

b. $P (5) = \frac{1}{10}$ $P(5)=1/10$

$SigmaP(5"x")=(1/10)*(10/10)=1/10*1=1/10$

$SigmaP(5"xx")=(1/10)*(10/10)*(10/10)=1/10*1*1=1/10$

$SigmaP(5"xxx")=(1/10)*(10/10)*(10/10)*(10/10)=1/10*1*1*1=1/10$

c. With one card: $P("No repeatition")=10/10=1$

With two cards: $P("No repeatition")=10/10*9/10=0.9$

Explanation: with one card, it is impossible to repeat numbers, so $P("No repeatition")=1$ . However, with two cards, there are 100 combinations, however, for every card, only 9 involve no repeatition. So, $P("No repeatition")=(9*10)/(10*10)=90/100=0.9$

With three cards: $P("No repeatition")=10/10*9/10*8/10=0.72$

With four cards: $P("No repeatition")=10/10*9/10*8/10*7/19=0.504$

d. $P("Not " 2, 3 or 4)=7/10$

$P(4("Not " 2, 3 or 4))=(7/10)^4=2401/10000$

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?Ten slips of paper are each marked with a different number (from 0 to 9) and placed in a basket. A slip is pulled out, its number is recored (in the order it is drawn), and the slip is replaced. This is done 4 times.

How to find the probabilities that the follow (numbers) are formed?

a) 1234
b) a number which starts with 5
c) a number with no repetition of digits.
d) a number that does not contain 2,3, or 4

2 Answers

Explanation:

Explanation:

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Impact of this question

Science

Math

Humanities

... and beyond

?Ten slips of paper are each marked with a different number (from 0 to 9) and placed in a basket. A slip is pulled out, its number is recored (in the order it is drawn), and the slip is replaced. This is done 4 times.

How to find the probabilities that the follow (numbers) are formed? a) 1234 b) a number which starts with 5 c) a number with no repetition of digits. d) a number that does not contain 2,3, or 4

2 Answers

Explanation:

Explanation:

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Impact of this question

How to find the probabilities that the follow (numbers) are formed?

a) 1234
b) a number which starts with 5
c) a number with no repetition of digits.
d) a number that does not contain 2,3, or 4