Tutorial on finding the probability of an event. In what follows, S is the sample space of the experiment in question and E is the event of interest. n(S) is the number of elements in the sample space S and n(E) is the number of elements in the event E.

Let us first write the sample space S of the experiment.

S = {1,2,3,4,5,6}

Let E be the event "an even number is obtained" and write it down.

E = {2,4,6}

We now use the formula of the classical probability.

P(E) = n(E) / n(S) = 3 / 6 = 1 / 2

The sample space S is given by.

S = {(H,T),(H,H),(T,H),(T,T)}

Let E be the event "two heads are obtained".

E = {(H,H)}

We use the formula of the classical probability.

P(E) = n(E) / n(S) = 1 / 4

a) -0.00001

b) 0.5

c) 1.001

d) 0

e) 1

f) 20%

A probability is always greater than or equal to 0 and less than or equal to 1, hence only

a) equal to 1

b) equal to 4

c) less than 13

a) The sample space S of two dice is shown below.

S = { (1,1),(1,2),(1,3),(1,4),(1,5),(1,6)

(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)

(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)

(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)

(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)

(6,1),(6,2),(6,3),(6,4),(6,5),(6,6) }

Let E be the event "sum equal to 1". There are no outcomes which correspond to a sum equal to 1, hence

P(E) = n(E) / n(S) = 0 / 36 = 0

b) Three possible outcomes give a sum equal to 4: E = {(1,3),(2,2),(3,1)}, hence.

P(E) = n(E) / n(S) = 3 / 36 = 1 / 12

c) All possible outcomes, E = S, give a sum less than 13, hence.

P(E) = n(E) / n(S) = 36 / 36 = 1

Let H be the head and T be the tail of the coin. The sample space S of the experiment described in question 5 is as follows

S = { (1,H),(2,H),(3,H),(4,H),(5,H),(6,H)

(1,T),(2,T),(3,T),(4,T),(5,T),(6,T)}

Let E be the event "the die shows an odd number and the coin shows a head". Event E may be described as follows

E={(1,H),(3,H),(5,H)}

The probability P(E) is given by

P(E) = n(E) / n(S) = 3 / 12 = 1 / 4

The sample space S of the experiment in question 6 is shwon below

Let E be the event "getting the 3 of diamond". An examination of the sample space shows that there is one "3 of diamond" so that n(E) = 1 and n(S) = 52. Hence the probability of event E occurring is given by

P(E) = 1 / 52

The sample space S of the experiment in question 7 is shwon above (see question 6)

Let E be the event "getting a Queen". An examination of the sample space shows that there are 4 "Queens" so that n(E) = 4 and n(S) = 52. Hence the probability of event E occurring is given by

P(E) = 4 / 52 = 1 / 13

We first construct a table of frequencies that gives the marbles color distributions as follows

color | frequency |
---|---|

red | 3 |

green | 7 |

white | 10 |

P(E) = Frequency for white color / Total frequencies in the above table

= 10 / 20 = 1 / 2

We construct a table of frequencies for the the blood groups as follows

group | frequency |
---|---|

a | 50 |

B | 65 |

O | 70 |

AB | 15 |

We use the empirical formula of the probability

P(E) = Frequency for O blood / Total frequencies

= 70 / 200 = 0.35

b) Two coins are tossed, find the probability that one

c) Two dice are rolled, find the probability that the sum is equal to 5.

d) A card is drawn at random from a deck of cards. Find the probability of getting the King of heart.

a) 2 / 6 = 1 / 3

b) 2 / 4 = 1 / 2

c) 4 / 36 = 1 / 9

d) 1 / 52

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