Q. 2.16

Question

Poker dice is played by simultaneously rolling 5 dice. 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}

Step-by-Step Solution

Verified
Answer

Hence proved.

1Step 1

Since we are rolling 5 dice, the total number of events is 65.

2Step 2 a) P{no two alike}

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

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

5! for rolling of 5 different dice.

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

Therefore, probability is 72065=.0962

3Step 3 b) P{one pair}

There are 61×52 ways 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 is 360065=.4630

4Step 4 c) P{two pair}

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

Thus, probability is 180065=.2315

5Step 5 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

6Step 6 e) P{full house}

Full house is three alike with pair

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

Thus, probability is 30065=.0386


7Step 7 f) P{four alike}

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

Thus, probability is 15065=.0193

8Step 8 g) P{five alike}

The number of five alike is 61=6.

Thus, probability is 665=.0008