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Sonja [21]
2 years ago
9

Starting with blue balls, in each of sequential rounds, we remove a random ball and replace it with a new red ball. for example,

after the first round we have 9 blue balls and one red ball, after the second round, with probability we have 8 blue balls and 2 red balls, and with probability we have 9 blue balls and one red ball, etc. what is the probability that the ball we remove at the 11th round is blue
Mathematics
1 answer:
KATRIN_1 [288]2 years ago
7 0

The probability that the ball we remove at the 11th round is blue will be 0.6.

<h3>How to calculate probability?</h3>

From the information given, we remove a random ball and replace it with a new red ball. for example, after the first round we have 9 blue balls and one red ball.

Since we have 9 blue balls and remove one every round. Let's assume that the number of blue and red balls will be 6 and 4 respectively in the 11 the round

Therefore, the probability that the ball we remove at the 11th round is blue will be:

= 6/(6 + 4)

= 6/10

= 0.6

Learn more about probability on:

brainly.com/question/24756209

#SPJ1

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For the writer, <span><span>there are 20
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2.       </span>2 + 38 = 40</span>
<span><span>3.       </span>3 + 37 = 40</span>
<span><span>4.       </span>4 + 36 = 40</span> <span><span>
5.       </span>5 + 35 = 40</span> <span><span>
6.       </span>6 + 34 = 40</span>
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<span><span>10.   </span>10 + 30 = 40</span> <span><span>
11.   </span>11 + 29 = 40</span> <span><span>
12.   </span>12 + 28 = 40</span>
<span><span>13.   </span>13 + 27 = 40</span> <span><span>
14.   </span>14 + 26 = 40</span> <span><span>
15.   </span>15 + 25 = 40</span> <span><span>
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17.   </span>17 + 23 = 40</span> <span><span>
18.   </span>18 + 22 = 40</span> <span><span>
19.   </span>19 + 21 = 40</span> <span><span>
20.   </span>20 + 20 = 40</span>



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