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The probability that a person in a particular evening class is left-handed is $\frac{1}{6}$. From a class of 15 women and 5 men a person is chosen at random. Assuming the 'left-handedness' is independent of the sex of a person, find the probability that the person chosen is a man or is left-handed.

Can anyone give some hints to me?

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    $\begingroup$ What is the expected number of right-handed women? $\endgroup$
    – Henry
    Jun 16, 2015 at 11:30
  • $\begingroup$ Hint: independent $\implies P(A\cap B)=P(A)P(B)$ $\endgroup$
    – danimal
    Jun 16, 2015 at 11:30
  • $\begingroup$ Hint: What is the definition of independent events? $\endgroup$ Jun 16, 2015 at 11:30

2 Answers 2

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$P(M\vee L)=1-P(M^c\wedge L^c)=\dots$ use independency here.

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$P$(man) OR $P$(left-handed)

$$=\dfrac5{15+5}+\dfrac16-\dfrac5{15+5}\cdot\dfrac16$$

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  • $\begingroup$ Should be 15+5 . $\endgroup$
    – Mathxx
    Jun 16, 2015 at 11:33
  • $\begingroup$ @Mathxx, Thanks, sorry for the typo $\endgroup$ Jun 16, 2015 at 11:33

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