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A four-digit number is formed by randomly selecting from the digits 0
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16 Apr 2022, 07:21
16 Apr 2022, 07:21
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Question Stats:
80% (02:14) correct
20% (01:26) wrong based on 10 sessions
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A four-digit number is formed by randomly selecting from the digits 0 through 9. No digit may be selected more than once. If the first two digits selected are 8 and 6, what are the chances that the number is even?
(A) 1/64
(B) 3/64
(C) 1/24
(D) 1/21
(E) 3/8
(A) 1/64
(B) 3/64
(C) 1/24
(D) 1/21
(E) 3/8
Part of the project: Skill Builder Project for a Q170
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
GRE Math Essentials for a TOP Quant Score - Q170!!! (2021)
GRE Geometry Formulas for a Q170
GRE - Math Book
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
GRE Math Essentials for a TOP Quant Score - Q170!!! (2021)
GRE Geometry Formulas for a Q170
GRE - Math Book
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A four-digit number is formed by randomly selecting from the digits 0
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29 Apr 2022, 02:17
29 Apr 2022, 02:17
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Carcass wrote:
A four-digit number is formed by randomly selecting from the digits 0 through 9. No digit may be selected more than once. If the first two digits selected are 8 and 6, what are the chances that the number is even?
(A) 1/64
(B) 3/64
(C) 1/24
(D) 1/21
(E) 3/8
(A) 1/64
(B) 3/64
(C) 1/24
(D) 1/21
(E) 3/8
Part of the project: Skill Builder Project for a Q170
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
GRE Math Essentials for a TOP Quant Score - Q170!!! (2021)
GRE Geometry Formulas for a Q170
GRE - Math Book
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
GRE Math Essentials for a TOP Quant Score - Q170!!! (2021)
GRE Geometry Formulas for a Q170
GRE - Math Book
The first and second place of a 4-digit number is fixed i.e. the number is 8 6 _ _
Now, let us fix the last places as 0, 2, and 4 and form cases;
Case I - 8 6 _ 0
we can plug 7 numbers for the thirs place (exclude 8, 6, and 0)
So, 7 numbers
Case II - 8 6 _ 2
we can plug 7 numbers for the thirs place (exclude 8, 6, and 2)
So, 7 numbers
Case III - 8 6 _ 4
we can plug 7 numbers for the thirs place (exclude 8, 6, and 4)
So, 7 numbers
Total available numbers = 7 + 7 + 7 = (3)(7)
Now, numbers of ways in which we can pick last 2 cards w/o repetition = (8)(7)
Thus, req. Prob. = \(\frac{(3)(7)}{(8)(7)} = \frac{3}{8}\)
Hence, option E
General Discussion
Re: A four-digit number is formed by randomly selecting from the digits 0
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06 Dec 2023, 01:48
06 Dec 2023, 01:48
1
Total Events = _ _ _ _ >> Total 10 number ( 0 - 9) two already taken for 1st and 2nd position thus 3rd position have 8 possibilities and 4th have 7 possibilities , thus total events = 8*7
Total events for even number >> even number from 0 - 9 i.e 0, 2, 4, 6, 8 out of which two already fixed at 1st and 2nd position, thus to make
it even number 4th position should contain even number , so 4th position have 3 possibilities and 3rd have 7 possibilities ( as out of 10 digit, 3 are fixed)
Total events for even number = 7*3
Probability = 7*3/8*7 = 3/8
Total events for even number >> even number from 0 - 9 i.e 0, 2, 4, 6, 8 out of which two already fixed at 1st and 2nd position, thus to make
it even number 4th position should contain even number , so 4th position have 3 possibilities and 3rd have 7 possibilities ( as out of 10 digit, 3 are fixed)
Total events for even number = 7*3
Probability = 7*3/8*7 = 3/8
