A die is tossed 6 times. Determine the probability for the number 4 to show up o

Question

A die is tossed 6 times. Determine the probability for the number 4 to show up on the face more than twice. 5 cards are drawn from a deck of 52 cards. Determine the possibility of getting a royal flush card from the drawing. Note that there are only 5 royal flush cards (hand with A, K, Q, J and 10) in the deck. At a certain event, 30 people attend but 5 of them will be chosen at random to receive door prizes. The prizes are all the same. Calculate how many different groups of five people that can be chosen to receive the prizes.

Explanation / Answer

a) P(number 4 more than twice) = P(number 4 three time) + P(number 4 four time) + P(number 5 three time) + P(number 6 three time)

= 1 – [ P(number 4 doesn’t occur) + P(number 4 one time) + P(number 4 two time) ]

This is a binomial distribution.

P(M) = NCM * PM *(1-P)N-M

Where N is total number of trials

P is the probability of success; P = 1/6

P(number 4 more than twice) = 1 – [6C0 * 1/60 *(1-1/6)6 + 6C1 * 1/61 *(1-1/6)5 + 6C2 * 1/62 *(1-1/6)4 ]

= 1 – [ 0.3348 + 0.4018 + 0.2009]

= 0.0625

b)The number of possible poker hands = 52C5 = 2598960

The best hand (because of the low probability that it will occur) is the royal flush, which consists of 10, J, Q, K, A of the same suit.

There are only 4 suits in a deck of 52 cards.

P(royal flush) = number of favourable outcome / total number of outcome

= 4/2598960 = 0.000001539

c) Total number of people = 30

Number of people to be chosen = 5

Number of possible groups = 30C5 = 30! /25! 5! = 142506

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