Answer to Question #130649 in Statistics and Probability for ajay

Question

Q1.Four cards are drawn at random from a pack of 52 cards. Find the probability that this consists
of:
a. a king, a queen, a jack and an ace.
b.two red and two blacks
c.two are kings and two are queens
d.there are two cards of hearts and two cards of diamonds

Q2.In shuffling a pack of cards four are accidentally dropped,find the chance that the missing cards should be one from each suit.

Q3.If the letters of the word regulations be arranged at random what is the probability that there will be exactly 4 letters between r and e.

Answer 1

ANSWER 1A)

The probability (P) of getting a king, a queen, a jack and an ace in a random selection of 4 cards from a pack of 52 cards is given by,

P = $\frac {number\ of\ favorable\ outcomes}{number\ of\ possible\ outcomes} = \frac{4C1*4C1*4C1*4C1}{52C4}$

P = $\frac {4*4*4*4}{\frac {52*51*50*49}{4*3*2*1}} = 0.0009456$

ANSWER 1B)

The probability (P) of getting 2 red and 2 black cards in a random selection of 4 cards from a pack of 52 cards is given by,

P = $\frac {number\ of\ favorable\ outcomes}{number\ of\ possible\ outcomes} = \frac{26C2*26C2}{52C4}$

P = $\frac {\frac {26*25*26*25}{2*1*2*1}}{\frac {52*51*50*49}{4*3*2*1}} =$ 0.3901

ANSWER 1C)

The probability (P) of getting 2 king and 2 queen cards in a random selection of 4 cards from a pack of 52 cards is given by,

P = $\frac {number\ of\ favorable\ outcomes}{number\ of\ possible\ outcomes} = \frac{4C2*4C2}{52C4}$

P = $\frac {\frac {4*3*4*3}{2*1*2*1}}{\frac {52*51*50*49}{4*3*2*1}}$ = 0.000133

ANSWER 1D)

The probability (P) of getting 2 heart and 2 diamond cards in a random selection of 4 cards from a pack of 52 cards is given by,

P = $\frac {number\ of\ favorable\ outcomes}{number\ of\ possible\ outcomes} = \frac{13C2*13C2}{52C4}$

P = $\frac {\frac {13*12*13*12}{2*1*2*1}}{\frac {52*51*50*49}{4*3*2*1}}$ = 0.02247

ANSWER 2)

The probability (P) of getting 1 card of each suit in a random selection of 4 cards from a pack of 52 cards is given by,

P = $\frac {number\ of\ favorable\ outcomes}{number\ of\ possible\ outcomes} = \frac{13C1*13C1*13C1*13C1}{52C4}$

P = $\frac {\frac {13*13*13*13}{1*1*1*1}}{\frac {52*51*50*49}{4*3*2*1}}$ = 0.1055

ANSWER 3)

Considering 1 group of 6 letters within the arrangement of 11 letter word where letter r'' & 'e' are at the extremes and the remaining 5 letters as 5 individual groups = total 6 groups.

So we can arrange these 6 groups (1 group of 6 letters and remaining 5 letters considered as individual groups) in 6P6 ways = 6! = 720 ways.

Now the letters 'r' & 'e' can be arranged such that 'r' is the 1st letter in the 5 letter group and 'e' is the last letter or vice-versa.

So 'r' & 'e' can be arranged in 2P2 ways = 2! = 2 ways within that group.

The remaining 4 letters in between 'r' & 'e' can be arranged in 4! = 24 ways.

The probability (P) of arranging the letters in the word "regulations" in such a away that there are exactly 4 letters between letter 'r' & letter 'e' is given by,

P = $\frac {number\ of\ favorable\ outcomes}{number\ of\ possible\ outcomes}$

P = $\frac {720*2*24}{11!} = 0.0008658$