A box contains 5 red, 7 blue and 4 green id..

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

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 - $\frac{37}{120}$ = $\frac{83}{120}$

$\frac{5}{16}$ * $\frac{4}{15}$ + $\frac{5}{16}$ * $\frac{7}{15}$ + $\frac{5}{16}$ * $\frac{4}{15}$

= $\frac{1}{12}$ + $\frac{7}{48}$ + $\frac{1}{12}$

$\frac{8 + 7}{48}$ = $\frac{15}{48}$

= $\frac{5}{16}$

Answer Details