Probability Part 1

Let S and NS be the respective events of choosing a spade card and a card which is not a spade.
P(S)=‌

‌13C1

‌52C1

=‌

1

4

P(NS)=‌

‌39C1

‌52C1

=‌

3

4

Let A be the event when both the chosen cards are spade.
Let A be the event when both the chosen cards are not spade.We have to find the probability of missing card not being spade when both the chosen cards are spade =P(S∣A)
By using Bayes theorem,
P(S∣A)=‌

P(S∩A)

P(A)

=‌

P(A∣S)P(S)

P(A∣S)P(S)+P(A∣S)P(S)

=‌

( ‌13C2 ‌51C2 )( ‌39C1 ‌52C1 )

(‌ ‌13C1 ‌52C1 )(‌ ‌12C2 ‌51C2 )+(‌ ‌13C2 ‌51C2 )(‌ ‌39C1 ‌52C1 )

=‌

( 13⋅12 51⋅50 )( 3 4 )

(‌ 1 4 )(‌ 12⋅11 51⋅50 )+(‌ 3 4 )(‌ 13⋅12 51⋅50 )

=‌

13×12×3

(12×11)+(3×12×13)

=‌

39

50