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20 Feb 2025, 01:30
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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)!
Difficulty:
35%
(medium)
Question Stats:
79% (02:12) correct
21%
(02:24)
wrong
based on 87
sessions
History
Date
Time
Result
Not Attempted Yet
A box contains 2 washers 3 nuts 4 bolts. the probability of drawing two washers first followed by 3 nuts and subsequently 4 bolts is
A. 1/346
B. 1/560
C. 1/1260
D. 1/5467
E. 1/5600
A. 1/346
B. 1/560
C. 1/1260
D. 1/5467
E. 1/5600
Updated on: 20 Feb 2025, 01:54
Originally posted by ManifestDreamMBA on 20 Feb 2025, 01:39.
Last edited by ManifestDreamMBA on 20 Feb 2025, 01:54, edited 1 time in total.
Last edited by ManifestDreamMBA on 20 Feb 2025, 01:54, edited 1 time in total.
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Let W be washer, N be nuts and B be bolts
Total outcomes = \(\frac{9!}{ 2!*3!*4!}\) = 1260
There's only 1 way to achieve WWNNNBBB
Probability = 1/1260
Total outcomes = \(\frac{9!}{ 2!*3!*4!}\) = 1260
There's only 1 way to achieve WWNNNBBB
Probability = 1/1260
Bunuel
20 Feb 2025, 01:52
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Bunuel
IMPORTANT POINT: The sequence of taking out the washer, nut and bolts is FIXED
Question: Probability (with fixed sequence) = ?
Given: 2 washers 3 nuts 4 bolts, i.e. Total count of items = 2+3+4 = 9
Probability of first washer = 2/9
Probability of Second washer = 1/8 (only 8 items are left)
Probability of first Nut = 3/7 (only 7 items are left of which 3 are nuts)
Probability of Second Nut = 2/6 (only 6 items are left)
Probability of Third Nut = 1/5 (only 5 items are left)
Probability of first bolt = 4/4 (only 4 items are left of which 3 are nuts)
Probability of Second bolt = 3/3 (only 3 items are left)
Probability of Third bolt = 2/2 (only 2 items are left)
Probability of Forth bolt = 1/1 (only 1 items are left)
Required Probability = (2/9)*(1/8)*(3/7)*(2/6)*(1/5)*(4/4)*(3/3)*(2/2)*(1/1) =1/1260
Answer: Option C
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