(ai) A bag contains 16 identical balls of which 4 are green. A boy picks a ball at random from the bag and replaces it. If this is repeated 5 times, what is...

(ai) \(p=\frac{4}{16}=\frac{1}{4}\)

\(∴q=1-\frac{1}{4}=\frac{3}{4}\)

P(The probability that he did not pick a green ball) = \(^nC_r p^rq^{n - r}\)

Where n = 5 and r = 0

\(=^5C_0(\frac{1}{4})^o(\frac{3}{4})^{5 - 0}\)

\(=^5C_0(\frac{1}{4})^o(\frac{3}{4})^5\)

\(=1\times1\times\frac{243}{1024}\)

\(=\frac{243}{1024}\)

(aii) Pr(at least three) = Pr(for 3 green balls) + Pr (for 4 green balls) + Pr (for 5 green balls)

\(^5C_3(\frac{1}{4})^3(\frac{3}{4})^{5 - 3}+^5C_4(\frac{1}{4})^4(\frac{3}{4})^{5 - 4}+^5C_5(\frac{1}{4})^5(\frac{3}{4})^{5 - 5}\)

\(=^5C_3(\frac{1}{4})^3(\frac{3}{4})^2+^5C_4(\frac{1}{4})^4(\frac{3}{4})^1+^5C_5(\frac{1}{4})^5(\frac{3}{4})^0\)

\(=10\times\frac{1}{6}4\times\frac{9}{16}+5\times\frac{1}{256}\times\frac{3}{4}+1\times\frac{1}{1024}\times1\)

\(=\frac{45}{512}+\frac{15}{1024}+\frac{1}{1024}\)

\(=\frac{53}{512}\)

(b) The sum of deviations from the mean is always equal to 0. This is a fundamental property of deviations and the definition of the mean.

\(= -2 + (m - 1) + (m^2 + 1) + (-1) + 2 + (2m - 1) + (-2) = 0\)

\(= -2 + m - 1 + m^2 + 1 - 1 + 2 + 2m - 1 - 2 = 0\)

\(= m^2 + 3m - 4 = 0\)

\(= m^2 + 4m - m - 4 = 0\)

= m(m + 4) - 1(m + 4) = 0

= (m + 4)(m - 1) = 0

This gives us two possible values for m: -4 and 1

So, the possible values of m are -4 and 1.