Probability Theory, Essay Example

Abstract

Probability theory is used in analysis of random phenomena. The chance that a specified event will occur is referred to as the probability. Its application is beneficial in instances when the outcome of a certain known event is not certain but can be estimated from the available information.

Introduction

The probability of an event defined by the above statement is equal to the number of designated options divided by total number of options that are available. Thus, the probability that one will draw a Dime is 2/11 which equals 0.181.On the other hand, the probability that the drawn piece is a Nickel=3/11, or 0.273. Similarly, the probability that the picker gets Quarter=6/11 or 0.546.
2. You are rolling a pair of dice, one red and one green. What is the probability of the following outcomes

The probability table depending on the number of dots is displayed as below. The table is drawn from the reference of one dice.

Total number of possible outcomes is 36. The probability that the sum of the two outcomes is 8(displayed by an asterisk)=5/36=0.14
b. The sum of the two numbers you roll is 12 (displayed by #) =1/36=0.03

c. The sum of the two numbers you roll is 7(displayed by @=) 6/36=0.17
3. For this question pretend you are drawing cards without replacement from the infamous “Iraq’s Most Wanted” deck issued by the U.S. Military before Saddam Hussein and his gang were killed or captured. If you are drawing from the full deck of 52 cards (no jokers), what are the following probabilities:
a. You draw a card that is not Saddam Hussein.

Since only one card is designated Saddam, thus the probability that the drawn card is not Saddam is 51/52=0.98
b. The probability that the card you draw first is Saddam is 1/52. The probability that the second drawn is Chemical Ali is 1/51. Thus, the probability that the first two cards drawn are designated Saddam and Ali is 1/52*1/51=0.000377
c. You draw 14 cards and not one of them is Saddam Hussein

For each card drawn, the denominator and numerator reduces by on since there is no replacement. The probability that the card at each draw for the fourteen cards is not designated Saddam is:

51/52*50/51*49/50*48/49*47/48*46/47*45/46*44/45*43/44*42/43*41/42*40/41*39/40*38/39=0.7152