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Nixondutta
Joined: 24 Apr 2017
Last visit: 10 Aug 2018
Posts: 40
Given Kudos: 83
Status:The journey is always more beautiful than the destination
Affiliations: Computer Science
Location: India
Concentration: Statistics, Strategy
Schools: Columbia EMBA"19 Simon '19 Mason '19
GMAT 1: 570 Q40 V28
GPA: 3.14
Updated on: 09 Apr 2018, 01:21
Originally posted by Nixondutta on 08 Apr 2018, 23:56.
Last edited by Bunuel on 09 Apr 2018, 01:21, edited 1 time in total.
Last edited by Bunuel on 09 Apr 2018, 01:21, edited 1 time in total.
Edited the question.
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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95% (00:37) correct
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A bag contains 7 red, 9 yellow, and 3 black balls. If a ball is picked at random, what is the probability that the ball drawn will be either a red or a black ball?
a. 7/19
b. 9/19
c. 10/19
d. 11/19
e. 12/19
Source- time4education
a. 7/19
b. 9/19
c. 10/19
d. 11/19
e. 12/19
Source- time4education
09 Apr 2018, 00:27
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Nixondutta
probablity=\(\frac{(7C1+3C1)}{(19C1)}\)
option C
04 Jun 2018, 13:55
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P(AorB) = P(A)+P(B)-P(A and B)
total balls =19
probablity of chooisng a red ball = 7/19 = P(A)
probablity of choosing a black ball = 3/19 = P(B)
probability of choosing a black and red ball = 0 (not possible as it either be red or black)
P(A or B) = 7/19 + 3/19 -0 = 10/19
so C
total balls =19
probablity of chooisng a red ball = 7/19 = P(A)
probablity of choosing a black ball = 3/19 = P(B)
probability of choosing a black and red ball = 0 (not possible as it either be red or black)
P(A or B) = 7/19 + 3/19 -0 = 10/19
so C
Nixondutta
