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Sagot :
The probability of selecting none of the correct six integers, when the order in which they are selected doesnot matter is 0.43.
According to thr question.
We have to find the probability of selecting none of the correct six integers from the positive integers not exceeding 48.
Let E be the event of selecting 6 numbers from 40 and S be the sample space of all integers not exceeding 48.
Now,
The total number of ways of selecting 6 numbers from 48
[tex]= ^{48} C_{6}[/tex]
[tex]= \frac{48!}{6!\times 42!}[/tex]
[tex]= \frac{48\times 47\times46\times45\times44\times43\times42!}{6!\times\ 42!}[/tex]
= 8835488640/6!
And, the total number of ways of selecting 6 incorrect numbers from 42
= [tex]^{42} C_{6}[/tex]
[tex]= \frac{42\times41\times40\times39\times38\times37\times36!}{6!\times36!}[/tex]
= 3776965920/6!
Therefore, the probability of selecting none of the correct six integers, when the order in which they are selected does not matter is given by
[tex]= \frac{^{42C_{6} } }{^{48} C_{6} }[/tex]
[tex]= \frac{\frac{3776965920}{6!} }{\frac{8835488640}{6!} }[/tex]
= 3776965920/8835488640
= 0.427
≈ 0.43
Hence, the probability of selecting none of the correct six integers, when the order in which they are selected doesnot matter is 0.43.
Find out more information about probability here:
https://brainly.com/question/17083464
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