A bag contains 4 red, 3 black and 5 blue balls. Find the probability of drawing
(i) a red ball
(ii) two blue balls
(iii) one red and one black ball
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Answered by
2
4 red balls, 3 black balls, 5 blue balls ;
Total number of outcomes
= Total number of balls
= 4+3+5
= 12
Probability = Number of favorable outcome/Total number of outcomes
(1) A Red Ball
Favorable outcomes = 4
Probability = 4/12 = (1/3)
(2) Two blue balls
Favorable outcomes = 5×2=10
Probability = 10/12 = (5/6)
(3) One red and one black ball
favorable outcomes = 4+3 = 7
Probability = 7/12
============================
Total number of outcomes
= Total number of balls
= 4+3+5
= 12
Probability = Number of favorable outcome/Total number of outcomes
(1) A Red Ball
Favorable outcomes = 4
Probability = 4/12 = (1/3)
(2) Two blue balls
Favorable outcomes = 5×2=10
Probability = 10/12 = (5/6)
(3) One red and one black ball
favorable outcomes = 4+3 = 7
Probability = 7/12
============================
Achal28300:
in the second part, it is two blue balls not one, so how is it 5/12?? Shouldn't it be (5/12)*2 ??
Right !
thanx for correcting me :)
Answered by
3
Total number of balls = 4+3+5 = 12
(i) Number of red balls = 4
Probability = 4/12 = 1/3
(ii) Number of two blue balls = 5 x 2 = 10
Probability = 10/12 = 5/6
(iii) Number of red and black balls = 4+ 3 = 7
Probability = 7/12
(i) Number of red balls = 4
Probability = 4/12 = 1/3
(ii) Number of two blue balls = 5 x 2 = 10
Probability = 10/12 = 5/6
(iii) Number of red and black balls = 4+ 3 = 7
Probability = 7/12
are you sure, it is right
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