Question

A random number generator draws at random with replacement from the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Find the chance that the digit 5 appears on more than 11% of the draws, if:

PROBLEM 2A : 1.0 POINTS

100 draws are made

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PROBLEM 2B : 1.0 POINTS

1000 draws are made

Answer 1

can any one reply pls

Answer 2

pls some one reply

Answer 3

ok i can give a shot but not so sure..
P(5)= (1/10)
for 100 draws
P(5 appears in 11% draw) =100C11* [(1/10)^11]*[(9/10)^(100-11)]

Answer 4

The probability that the digit 5 appears on more than 11% of the draws if 1000 draws are made can be found from the normal approximation to the binomial distribution.
The mean = np = 1000 * 1/10 = 100
The standard distribution is given by
\[\sigma=\sqrt{np(1-p)}=\sqrt{90}=9.4868\]
The z-score for 110 draws is
\[z=\frac{X-\mu}{\sigma}=\frac{110-100}{9.4868}=1.054\]
Now you need to refer to a standard normal distribution table to find the cumulative probability corresponding to z = 1.054. This will give the cumulative probability that the digit 5 will appear up to 110 times in 1000 draws. To find probability that the digit 5 appears more than 110 times in 1000 subtract the probability just found from 1.0000.

Answer 5

The same method can be used to find the probability that the digit 5 appears on more than 11% of the draws if 100 draws are made.

Answer 6

@dinakar Do you understand the method?

Answer 7

pls can u reply the exact ans .i dnt have any idea abt ths method .i dint go through ecxrcies

Answer 8

On OS we try to help students to get solutions rather than give answers. However I will help you further with the 1000 draws question.
A standard normal distribution table gives the cumulative probability of 0.8540 for a z-score of 1.054.
The required probability is 1.0000 - 0.8540 = 0.146 or 14.6%

Answer 9

ans 14.6% went wrong

Answer 10

1)0.297
2)0.135

Answer 11

The accuracy of the approximation can be improved by applying a 'continuity correction' of 0.5 to the calculation of the z-score. The z-score then becomes
\[z=\frac{110+0.5-100}{9.4868}=1.107\]
The required probability is 1.0000 - 0.8786 = 0.1214 or 12.14%

Answer 12

thanks fr reply

Answer 13

A die is rolled 12 times. Find the chance that the face with six spots appears once among the first 6 rolls, and once among the next 6 rolls.

Answer 14

can any one post ans

Answer 15

I toss a coin 4 times. Find the chance of getting 2heads

Answer 16

actualy i have tried it ,i got 0.25 ans but it went wrong .pls any one can xplain

Answer 17

ans;p(HH)=1/2*1/2=0.25.but it went wrong .pls help me out

Answer 18

plssss someone help me out yar