In a game of chance, six cards numbered 1 to 6 are lying face down on a table. To cards are selected without replacement and the sum of both numbers is noted. Find the probability that the total sum is at least 10?

2 Answers
G_Ozdilek
Oct 24, 2017

Total probability vs. desired outcome

Explanation:

Your changes (1;1), (1;2), (1;3), (1;4), (1;5), (1;6), (2;1), (2;2), (2;3), (2;4), (2;5), (2;6), (3;1), (3;2), (3;3), (3;4), (3;5), (3;6), (4;1), (4;2), (4;3), (4;4), (4;5), (4;6),(5;1), (5;2), (5;3), (5;4), (5;5), (5;6), (6;1), (6;2), (6;3), (6;4), (6;5), (6;6).

The desired outcome: (4;6), (5;5), (5;6), (6;4), (6;5), and (6;6).

Therefore P(outcome at least 10)=6.

P(total outcome)=36.

Your probability = 6/36 =1/6

Geoff K.
Oct 29, 2017

The probability is #2/15#.

Explanation:

Since we are sampling without replacement, the possible pairs we could select are:

#S={color(white)[ color(black)((12,13,14,15,16),(,23,24,25,26),(,,34,35,36),(,,,45,46),(,,,,56))]}#

each with equal probability #1/absS=1/15.#

Let #E# be the event that the two cards have a sum of at least 10. The elements of #S# that correspond to event #E# are #(4,6)# and #(5,6),# meaning that the probability of drawing a sum of at least 10 in two cards is

#P("sum of two cards is at least 10")#

#="number of favourable outcomes"/"number of total outcomes"#

#=absE/absS=2/15.#

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