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Charra [1.4K]
3 years ago
12

1. In a table of random digits, Each digit is to occur with a probability of 0.1

Mathematics
1 answer:
anyanavicka [17]3 years ago
8 0
(A) Just because every digit has an equal chance of appearing does not mean that all will be equally represented. (See "gambler's fallacy")

(B) The experimental procedure isn't exactly clear, so assuming a table of digits refers to a table of just one-digit numbers, each with 0.1 chance of appearing (which means you can think of the digits 0-9), you should expect any given digit to appear about 0.1 or 10% of the time.

So if a table consists of 1000 digits, one could expect 7 to appear in 10% of the table, or about 100 times.
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What is the solution for this system of equations? <br>y= -1/4x + 6<br>y= x+1​
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  (x, y) = (4, 5)

Step-by-step explanation:

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A bag contain 5 red marbles,4 green marbles and 1 blue marble. A marble is chosen at random from the bag and not replaced; then
jarptica [38.1K]

Answer:

\frac{2}{15}

Step-by-step explanation:

GIVEN: A bag contain 5 red marbles,4 green marbles and 1 blue marble. A marble is chosen at random from the bag and not replaced, then a second marble is chose.

TO FIND: What is the probability both marbles are green.

SOLUTION:

Total marbles in bag =10

total number of green marbles =4

Probability that first marble will be green P(A)=\frac{\text{total green marbles}}{\text{total marbles}}

                                                                    =\frac{4}{10}=\frac{2}{5}

Probability that second marble will be green P(B)=\frac{\text{total green marble in bag}}{\text{total marble in bag}}

                                                                                    \frac{3}{9}=\frac{1}{3}

As both events are disjoint

probability both marbles are green =P(A)\times P(B)

                                                           =\frac{2}{5}\times\frac{1}{3}=\frac{2}{15}

Hence  the probability both marbles are green is \frac{2}{15}

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