Poker dice is played by simultaneously rolling 5dice. Show that

(a) P{no two alike}

(b) P{one pair}

(c) P{two pair}

(d) P{three alike}

(e) P{full house}

(f) P{four alike}

(g) P{five alike}

Short Answer

Expert verified

Hence proved.

Step by step solution

01

Step 1

Since we are rolling 5 dice, the total number of events is65.

02

a) P{no two alike}

No two alike i.e. we want 5different numbers from 5dice.

There are 65ways to choose 5different numbers from 1to 6.

5!for rolling of 5different dice.

Thus, number of no two alike is 5!×65=720

Therefore, probability is 72065=.0962

03

b) P{one pair}

There are 61×52ways to choose a pair and 53×3!ways to choose which is not a pair.

Thus, number of one pair is61×52×53×3!=3600.

Therefore, probability is360065=.4630

04

c) P{two pair}

The number of two pairs is 62×52×32×41=1800.

Thus, probability is180065=.2315

05

d) P{three alike}

First choose common number and dice having common number then choose other two numbers from five choisces.

The number of three alike is 61×53×52×2!=1200.

Thus, probability is 120065=.1543

06

e) P{full house}

Full house is three alike with pair

The number of full house is 61×53×51=300.

Thus, probability is30065=.0386

07

f) P{four alike}

The number of four alike is 61×54×51=150.

Thus, probability is 15065=.0193

08

g) P{five alike}

The number of five alike is 61=6.

Thus, probability is665=.0008

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A group of 6men and 6women is randomly divided into 2groups of size 6each. What is the probability that both groups will have the same number of men?

For any sequence of events E1,E2,...,define a new sequenceF1,F2,...of disjoint events (that is, events such that FiFj=Øwhenever ij) such that for alln1, 1nFi=1nEi

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