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

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

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

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...

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 =\(\frac{6}{35}\)

Hi Bunuel, what if we wanted to find the probability of not getting duplicate letters, and subtract this by 1?

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

Hi Rich ty for replying.

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?