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.
Calculate the probability of picking: (a) same color of bulbs; (6) different color of bulbs (c) at least one red bulb.
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Correct Answer: Option
Explanation:
. Total bulbs: n(Red)=5, n(B) = 7, n(G) = 4
(5+7+4) = 16
p(R)= 5/16, p(B) = 7/16, p(G) = 4/16
(a) p(All same colour of bulbs) = p(RR) Or p(BB) or p(GG)
\(\frac{5}{16}\) * \(\frac{4}{15}\) + \(\frac{7}{16}\) * \(\frac{6}{15}\) + \(\frac{4}{16}\) * \(\frac{3}{15}\)
= \(\frac{1}{12}\) + \(\frac{7}{40}\) + \(\frac{1}{20}\)
= \(\frac{10+21+6}{120}\) = \(\frac{37}{120}\)
(b) All different colors = 1-p (AIl the same colour) =
1 - \(\frac{37}{120}\) = \(\frac{83}{120}\)
(c) p(At least one red) = p(RR) + p(RB) + p(RG)
\(\frac{5}{16}\) * \(\frac{4}{15}\) + \(\frac{5}{16}\) * \(\frac{7}{15}\) + \(\frac{5}{16}\) * \(\frac{4}{15}\)
= \(\frac{1}{12}\) + \(\frac{7}{48}\) + \(\frac{1}{12}\)
= \(\frac{8+7}{48}\) = \(\frac{15}{48}\)
= \(\frac{5}{16}\)
. Total bulbs: n(Red)=5, n(B) = 7, n(G) = 4
(5+7+4) = 16
p(R)= 5/16, p(B) = 7/16, p(G) = 4/16
(a) p(All same colour of bulbs) = p(RR) Or p(BB) or p(GG)
\(\frac{5}{16}\) * \(\frac{4}{15}\) + \(\frac{7}{16}\) * \(\frac{6}{15}\) + \(\frac{4}{16}\) * \(\frac{3}{15}\)
= \(\frac{1}{12}\) + \(\frac{7}{40}\) + \(\frac{1}{20}\)
= \(\frac{10+21+6}{120}\) = \(\frac{37}{120}\)
(b) All different colors = 1-p (AIl the same colour) =
1 - \(\frac{37}{120}\) = \(\frac{83}{120}\)
(c) p(At least one red) = p(RR) + p(RB) + p(RG)
\(\frac{5}{16}\) * \(\frac{4}{15}\) + \(\frac{5}{16}\) * \(\frac{7}{15}\) + \(\frac{5}{16}\) * \(\frac{4}{15}\)
= \(\frac{1}{12}\) + \(\frac{7}{48}\) + \(\frac{1}{12}\)
= \(\frac{8+7}{48}\) = \(\frac{15}{48}\)
= \(\frac{5}{16}\)