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June 2021 p51 q7
2890
Sharma knows that she has 3 tins of carrots, 2 tins of peas and 2 tins of sweetcorn in her cupboard. All the tins are the same shape and size, but the labels have all been removed, so Sharma does not know what each tin contains.
Sharma wants carrots for her meal, and she starts opening the tins one at a time, chosen randomly, until she opens a tin of carrots. The random variable \(X\) is the number of tins that she needs to open.
(a) Show that \(P(X = 3) = \frac{6}{35}\).
(b) Draw up the probability distribution table for \(X\).
(c) Find \(\text{Var}(X)\).
Solution
(a) To find \(P(X = 3)\), we calculate the probability of opening two non-carrot tins followed by a carrot tin. The probability of the first tin not being carrots is \(\frac{4}{7}\) (since there are 4 non-carrot tins out of 7 total). The probability of the second tin not being carrots is \(\frac{3}{6}\) (since there are now 3 non-carrot tins out of 6 remaining). The probability of the third tin being carrots is \(\frac{3}{5}\) (since there are 3 carrot tins out of 5 remaining). Therefore, \(P(X = 3) = \frac{4}{7} \times \frac{3}{6} \times \frac{3}{5} = \frac{6}{35}\).
(b) The probability distribution table for \(X\) is:
\(x\)
1
2
3
4
5
\(p\)
\(\frac{15}{35}\)
\(\frac{10}{35}\)
\(\frac{6}{35}\)
\(\frac{3}{35}\)
\(\frac{1}{35}\)
(c) To find \(\text{Var}(X)\), we first calculate \(E(X)\):
