An-irregular-6-faced-die-is-thrown-and-the-expectation-that-in-10-throws-it-will-give-five-even-numbers-is-twice-the-expectation-that-it-will-give-four-even-numbers-How-many-times-in-15000-sets-of-10-

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Question Number 57952 by necx1 last updated on 14/Apr/19

An irregular 6 faced die is thrown and the expectation that in 10 throws it will give five even numbers is twice the expectation that it will give four even numbers.How many times in 15000 sets of 10 throws would you expect it to give one even number?

A n i r r e g u l a r 6 f a c e d d i e i s t h r o w n a n d

t h e e x p e c t a t i o n t h a t i n 10 t h r o w s i t w i l l

g i v e f i v e e v e n n u m b e r s i s t w i c e t h e

e x p e c t a t i o n t h a t i t w i l l g i v e f o u r e v e n

n u m b e r s . H o w m a n y t i m e s i n 15000

s e t s o f 10 t h r o w s w o u l d y o u e x p e c t i t

t o g i v e o n e e v e n n u m b e r ?

Answered by MJS last updated on 15/Apr/19

((p^5 (1−p)^5 )/(p^4 (1−p)^6 ))=2 (p/(1−p))=2 ⇒ p=(2/3) p(1−p)^9 (((10)),(( 9)) )=(2/3)×(1/3^9 )×10=((20)/(59049)) 15000×((20)/(59049))=((100000)/(19683))≈5

\frac{p^{5} {(1 - p)}^{5}}{p^{4} {(1 - p)}^{6}} = 2

\frac{p}{1 - p} = 2 \Rightarrow p = \frac{2}{3}

p {(1 - p)}^{9} (\begin{matrix} 10 \\ 9 \end{matrix}) = \frac{2}{3} \times \frac{1}{3^{9}} \times 10 = \frac{20}{59049}

15000 \times \frac{20}{59049} = \frac{100000}{19683} \approx 5

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An-irregular-6-faced-die-is-thrown-and-the-expectation-that-in-10-throws-it-will-give-five-even-numbers-is-twice-the-expectation-that-it-will-give-four-even-numbers-How-many-times-in-15000-sets-of-10-

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