A simple random sample of size five is selected from 52 card

A simple random sample of size five is selected from 52 cards. Find the probability of each of the following events.

(a) The five-card hand contains all four 2\'s.

(b) The five-card hand contains two aces and two 7\'s.

(c) The five-card hand contains three kings and the other two cards are of different denomination.

Solution

we can select 5 cards from 52 cards in ⁵²C₅ ways. so number of all possible cases is ⁵²C₅

a) let A be the event that the hand contains all four 2\'s.

from 4 2\'s we can select all 4 in ⁴C₄ ways. other one can be selected from remaining 48 cards in ⁴⁸C₁ ways

so P[A]=⁴C₄*⁴⁸C₁/ ⁵²C₅=0.0000184 [answer]

b) let B be the event that the hand contains two aces and two 7\'s.

from 4 aces we can select 2 aces in ⁴C₂ ways and from 4 7\'s we can select 2 7\'s in ⁴C₂ ways. the remaining one card can be chosen from 52cards-4 aces-4 7\'s=44 cards ins ⁴⁴C₁ ways.

so P[B]=⁴C₂*⁴C₂*⁴⁴C₁/⁵²C₅=0.000609 [answer]

c) let C be the event that the hand contains 3 kings and two other are of different denomination.

so out of 4 kings 3 can be chosen in ⁴C₃ ways. the other two cards are not kings and are of different denomination.

so there are 48 cards without the king.we can select one from them in ⁴⁸C₁ ways. the chosen card is of a denomination. so excluding that denomination and kings there are 44 cards. we can select one from them in ⁴⁴C₁ ways.

so P[C]=⁴C₃*⁴⁸C₁*⁴⁴C₁/⁵²C₅=0.00325 [answer]