A bag contains 6 red and 4 white balls. Two balls are drawn from the bag randomly. What is the probability that they are of the; (i) same colour (ii) different colour?
Answers
Answer:6/10 = 3/5
Step-by-step explanation:
Answer:
answer
Given a bag contains 66 white and 44 red balls.
\therefore∴ The sample space is S=6+4=10S=6+4=10
One ball is red and two balls are white.
CaseI:I: If red ball is drawn at first position and then the white balls
P=\dfrac{6}{10}\times\dfrac{4}{9}\times\dfrac{3}{8}=\dfrac{6\times 4\times 3}{10\times 9\times 8}=\dfrac{1}{10}P=
10
6
×
9
4
×
8
3
=
10×9×8
6×4×3
=
10
1
CaseII:II: If red ball is drawn at second position and then the white balls correspondingly
P=\dfrac{4}{10}\times\dfrac{6}{9}\times\dfrac{3}{8}=\dfrac{4\times 6\times 3}{10\times 9\times 8}=\dfrac{1}{10}P=
10
4
×
9
6
×
8
3
=
10×9×8
4×6×3
=
10
1
CaseIII:III: If red ball is drawn at third position
P=\dfrac{4}{10}\times\dfrac{3}{9}\times\dfrac{6}{8}=\dfrac{4\times 3\times 6}{10\times 9\times 8}=\dfrac{1}{10}P=
10
4
×
9
3
×
8
6
=
10×9×8
4×3×6
=
10
1
\therefore∴ Probability of one ball is red and two balls are white=\dfrac{1}{10}+\dfrac{1}{10}+\dfrac{1}{10}=\dfrac{3}{10}=
10
1
+
10
1
+
10
1
=
10
3
Step-by-step explanation:
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