Revision as of 02:59, 25 June 2024 by Admin (Created page with "'''Solution: C''' Let <math>N</math> be the number of boys. Clearly <math>N \leq 1 </math>. If <math>N=0 </math> then we have three girls and the probability of this event is (1/2)<sup>3</sup>. Hence the probability distribution for <math>N</math> is <math>P(N=0) = 0.125, P(N=1) = 0.875 </math>. Then <math>E[N] = 0.875 </math> and <math>E[N^2] = E[N] = 0.875 </math> and the variance equals 0.875 - 0.875<sup>2</sup> = 0.109375.")
Exercise
ABy Admin
Jun 25'24
Answer
Solution: C
Let [math]N[/math] be the number of boys. Clearly [math]N \leq 1 [/math]. If [math]N=0 [/math] then we have three girls and the probability of this event is (1/2)3. Hence the probability distribution for [math]N[/math] is [math]P(N=0) = 0.125, P(N=1) = 0.875 [/math]. Then [math]E[N] = 0.875 [/math] and [math]E[N^2] = E[N] = 0.875 [/math] and the variance equals 0.875 - 0.8752 = 0.109375.