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# There are two names given JOHNSON and JONES. If one letter is picked

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Joined: 03 Sep 2014
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There are two names given JOHNSON and JONES. If one letter is picked  [#permalink]

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Updated on: 12 Oct 2014, 01:29
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13
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Difficulty:

65% (hard)

Question Stats:

60% (02:15) correct 40% (02:18) wrong based on 146 sessions

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There are two names given JOHNSON and JONES. If one letter is picked from both simultaneously at random, then find the probability that the letter is same?

A. 24/35
B. 17/35
C. 8/25
D. 6/35
E. 1/25

Originally posted by eggplantpower on 11 Oct 2014, 18:27.
Last edited by Bunuel on 12 Oct 2014, 01:29, edited 1 time in total.
Renamed the topic.
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Re: There are two names given JOHNSON and JONES. If one letter is picked  [#permalink]

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12 Oct 2014, 01:38
4
2
eggplantpower wrote:
There are two names given JOHNSON and JONES. If one letter is picked from both simultaneously at random, then find the probability that the letter is same?

A. 24/35
B. 17/35
C. 8/25
D. 6/35
E. 1/25

JOHNSON
JONES

The probability that letter J is picked = 1/7*1/5 = 1/35;
The probability that letter O is picked = 2/7*1/5 = 2/35;
The probability that letter N is picked = 2/7*1/5 = 2/35;
The probability that letter S is picked = 1/7*1/5 = 1/35.

The sum of these probabilities = 6/35.

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Re: There are two names given JOHNSON and JONES. If one letter is picked  [#permalink]

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12 Oct 2014, 00:11
eggplantpower wrote:
There are two names given JOHNSON and JONES. If one letter is picked from both simultaneously at random, then find the probability that the letter is same?

A) 24/35
B) 17/35
C) 8/25
D) 6/35
E) 1/25

Can someone explain if the answer D has a typo? i am pretty sure answer is 4/35 but that isnt one of the choices...
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Re: There are two names given JOHNSON and JONES. If one letter is picked  [#permalink]

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12 Oct 2014, 01:46
JOHNSON - 7 letters
JONES - 5 letters

Common Letters : J, O, N, S

Prob of picking J from both = $$\frac{1}{7}+\frac{1}{5}=\frac{1}{35}$$
Prob of O = $$\frac{2}{7}+\frac{1}{5}=\frac{2}{35}$$
Prob of N = $$\frac{2}{7}+\frac{1}{5}=\frac{2}{35}$$
Prob of S = $$\frac{1}{7}+\frac{1}{5}=\frac{1}{35}$$

Total prob =$$\frac{6}{35}$$
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Re: There are two names given JOHNSON and JONES. If one letter is picked  [#permalink]

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18 Dec 2015, 09:47
1
Hi All,

This question can also be solved by using 'brute force' - you just have to 'map out' the possibilities.

The name "Johnson" has 7 letters and the name "Jones" has 5 letters, so there are (7)(5) = 35 possible pairings of letters that can occur.

Of those 35, the pairings that include the SAME letter both times are:

J and J
1st O and O
1st N and N
S and S
2nd O and O
2nd N and N

Total: 6

So the probability that the two randomly selected letters will match is 6/35.

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Re: There are two names given JOHNSON and JONES. If one letter is picked  [#permalink]

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13 Mar 2020, 11:14
Bunuel wrote:
eggplantpower wrote:
There are two names given JOHNSON and JONES. If one letter is picked from both simultaneously at random, then find the probability that the letter is same?

A. 24/35
B. 17/35
C. 8/25
D. 6/35
E. 1/25

JOHNSON
JONES

The probability that letter J is picked = 1/7*1/5 = 1/35;
The probability that letter O is picked = 2/7*1/5 = 2/35;
The probability that letter N is picked = 2/7*1/5 = 2/35;
The probability that letter S is picked = 1/7*1/5 = 1/35.

The sum of these probabilities = 6/35.

P.S. Please read carefully and follow: http://gmatclub.com/forum/rules-for-pos ... 33935.html Pay attention to rules 3 and 10. Also, please tag questions properly! This is NOT your question!

Hi Bunuel, what if we wanted to find the probability of not getting duplicate letters, and subtract this by 1?
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Re: There are two names given JOHNSON and JONES. If one letter is picked  [#permalink]

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13 Mar 2020, 13:37
Hi exc4libur,

You can certainly calculate the probability of NOT getting a matching pair of letters (and then subtract that fraction from 1), but that process would take a bit more work though.

You'll notice that Bunuel's work required 4 individual calculations which were then summed. Your approach would require 6 individual calculations, which are then summed and then that sum would need to be subtracted from 1.

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There are two names given JOHNSON and JONES. If one letter is picked  [#permalink]

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14 Mar 2020, 06:52
EMPOWERgmatRichC wrote:
Hi exc4libur,

You can certainly calculate the probability of NOT getting a matching pair of letters (and then subtract that fraction from 1), but that process would take a bit more work though.

You'll notice that Bunuel's work required 4 individual calculations which were then summed. Your approach would require 6 individual calculations, which are then summed and then that sum would need to be subtracted from 1.

GMAT assassins aren't born, they're made,
Rich

I understand the gmat requires us to master the most efficient way to solve problems.

Nonetheless, could you show me how you would calculate the probability of NOT getting the matching pairs, so I can see where I went wrong in my calculations?
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Re: There are two names given JOHNSON and JONES. If one letter is picked  [#permalink]

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14 Mar 2020, 12:37
1
Hi exc4libur,

Working with the work JOHNSON and JONES, here are the probabilities for pairs of letters that do NOT match (the first letter in each pair comes from the word JOHNSON and the second comes from JONES):

(J)(not J) = (1/7)(4/5) = 4/35
(O)(not O) = (1/7)(4/5) = 4/35
(H)(not H) = (1/7)(5/5) = 5/35
(N)(not N) = (1/7)(4/5) = 4/35
(S)(not S) = (1/7)(4/5) = 4/35
(O)(not O) = (1/7)(4/5) = 4/35
(N)(not N) = (1/7)(4/5) = 4/35

Total = (6)(4/35) + (5/35) = 29/35 --> again, this is the probability of NOT having a matching pair of letters. You could technically 'combine' the two "O" calculations into one calculation and the two "N" calculations into one calculation (but the overall result would be the same).

Probability of having a matching pair of letters = 1 - 29/35 = 6/35

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Re: There are two names given JOHNSON and JONES. If one letter is picked   [#permalink] 14 Mar 2020, 12:37