Question

Two cards are randomly selected from a deck of 52 playing cards. (a) What is the probability they constitute a pair (that is, that they are of the same denomination)? (b) What is the conditional probability they constitute a pair given that they are of different suits?

Answer 1

a)

For first card

favourable ways = selecting 1 card from 52 cards =52C1 =52

total ways =selecting 1 card from 52 cards =52C1 =52

so prob =52/52 =1

For second card

there are 4 alike cards in a deck,so to get a like pair,we have to select same denominator card from the rest of 3 cards,so this can be done is 3C1 ways =3

number of cards left are 51

so total ways =selecting 1 card from 51 cards =51C1 =51

so prob =3/51

Hence,prob(pair) =(1)*(3/51)

                              =3/51

                              =1/17

b)

For first card

favourable ways = selecting 1 card from 52 cards =52C1 =52

total ways =selecting 1 card from 52 cards =52C1 =52

so prob =52/52 =1

For second card

there are 4 alike cards in a deck,so to get a like pair,we have to select same denominator card from the rest of 3 cards,so this can be done is 3C1 ways =3

number of cards left of different suit is (52-13) =39

so total ways =selecting 1 card from 39 cards =39C1 =39

so prob =3/39

Hence,prob(pair) =(1)*(3/39)

                              =3/39

                              =1/13