From first 100 natural numbers, 3 numbers are selected. If these numbers are in A.P., then the probability that these numbers are even is:
From first 100 natural numbers, 3 numbers are selected. If these numbers are in A.P., then the probability that these numbers are even is:
Option 1 - <p>2/9<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
Option 2 - <p>29/56<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
Option 3 - <p>2/7</p>
Option 4 - <p>24/49</p>
4 Views|Posted 5 months ago
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1 Answer
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Answered by
5 months ago
Correct Option - 3
Detailed Solution:
A: 3 numbers are in A.P.
B: 3 numbers are even
P (B/A) = n (A ∩ B) / n (A) = (²? C? + ²? C? ) / (? C? +? C? ) = (25 * 24) / (50 * 49) = 12/49
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