(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 the probability that he:
did not pick a green ball;
(aii) 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 the probability that he:
picked a green ball at least three times?
(b) The deviations from a mean of values from a set of data are \(-2, ( m – 1), ( m ^2 + 1), -1, 2, (2 m – 1)\) and \(-2\). Find the possible values of \(m\) .
Explanation
(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.