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Please someone help is required to solve this out
6 (i) A manufacturer of biscuits produces 3 times as many cream ones as chocolate ones. Biscuits are
chosen randomly and packed into boxes of 10. Find the probability that a box contains equal
numbers of cream biscuits and chocolate biscuits.
(ii) A random sample of 8 boxes is taken. Find the probability that exactly 1 of them contains equal
numbers of cream biscuits and chocolate biscuits.
(iii) A large box of randomly chosen biscuits contains 120 biscuits. Using a suitable approximation,
find the probability that it contains fewer than 35 chocolate biscuits.
This is from Nov 2002 P6 (S1)
Thanks.
3 times as many cream ones as chocolate ones therefore if chocolate is x then cream will be 3x so x+3x=1, 4x=1, x=1/4 and 3x =3/4
Using Binomial we get 10C5*(0.25)^5*(0.75)^5=0.0584
8C1*(0.0584)(1-0.0584)^7=0.307
mean=npq=120*0.25=30
variance=npq=120*0.25*0.75=22.5
P(X<35)~~~~ continuity correction we get P(X<34.5)
P(Z<(34.5-35)/sqr22.5))=P(Z<0.949)
=0.829