A box contains 5 red, 7 blue and 4 green identical bulbs. Two bulbs are picked at random from the box without replacement. Calculate the probability of pick...
Question 1 Report
A box contains 5 red, 7 blue and 4 green identical bulbs. Two bulbs are picked at random from the box without replacement.
Calculate the probability of picking:
(a) same color of bulbs; (6) different color of bulbs (c) at least one red bulb.
Answer Details
The box holds \(5\) red, \(7\) blue and \(4\) green bulbs, giving \(5+7+4=16\) bulbs. Two bulbs are drawn without replacement, so the number of equally likely selections is
\[\binom{16}{2}=\frac{16\times 15}{2}=120.\]
(a) Same colour
Count the same-colour pairs for each colour and add:
- Both red: \(\binom{5}{2}=10\)
- Both blue: \(\binom{7}{2}=21\)
- Both green: \(\binom{4}{2}=6\)
Favourable outcomes \(=10+21+6=37\), so
\[P(\text{same colour})=\frac{37}{120}.\]
(b) Different colours
Every draw is either the two bulbs matching or not, so "different colours" is the complement of "same colour":
\[P(\text{different})=1-\frac{37}{120}=\frac{83}{120}.\]
(c) At least one red
Use the complement "no red bulb". There are \(16-5=11\) non-red bulbs, giving \(\binom{11}{2}=\frac{11\times 10}{2}=55\) all-non-red pairs. Hence
\[P(\text{no red})=\frac{55}{120}=\frac{11}{24},\qquad P(\text{at least one red})=1-\frac{11}{24}=\frac{13}{24}.\]
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