What is the probability that a) A card chosen from an ordinary deck of 52 cards is an ace? b) A randomly selected integer chosen from the first 100 positive integers is odd? c) A randomly selected day of the year 2020 is in May? d) The sum of the numbers on two dice is even when they are rolled? e) A fair coin lands Heads 6 times in a row? f) A fair coin lands Heads 4 times out of 5 flips? g) The bit string has exactly two 1s, given that the string begins with a 1 if you pick a bit string from the set of all bit strings of length ten? h) The bit string has the sum of its digits equal to seven if you pick a bit string from the set of all bit strings of length ten?

Respuesta :

Answer:

a) 0.076

b) 0.5

c) 0.084

d) 0.25

e) 0.00024

f) 0.156

Step-by-step explanation:

a) In an ordinary deck there are 4 aces. The probability of getting an ace is given by:

[tex]P=\frac{n}{N}[/tex]

n: options for an ace = 4

N: total number of cards = 52

[tex]P=\frac{4}{52}=0.076[/tex]

The probability is 7.6%

b) On the first 100 positive integers, there are 50 odd integer numbers. Then you have:

[tex]P=\frac{50}{100}=0.5[/tex]

The probability of getting an odd number is 50%

c) From a total of 365 day in the year, May has 31 day. Then you have:

[tex]P=\frac{31}{365}=0.084[/tex]

The probability is 8.4%

d) In this case you have two events. To get an odd number with the sum of the number of two dices you need and odd number in one dice and a pair number in the other dice.

The probability is the product of the occurrence of both events. The probability for an odd number is 1/2 and for a pair number is 1/2.

[tex]P=P_{odd}P_{pair}=(\frac{1}{2})(\frac{1}{2})=\frac{1}{4}=0.25[/tex]

The probability is 25%

e) To obtain Head 6 times in a row, the probability is:

[tex]P=\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}=0.00024[/tex]

Probability of 0.024%

f) To get Head 4 times from 5 flights, you first calculate the probability for 1 head in five flips:

[tex]P=(\frac{1}{2})^5=\frac{1}{32}[/tex]

There are 5 forms of getting 4 Heads for 5 flips, then you have:

[tex]P=5(\frac{1}{32})=\frac{5}{32}=0.156[/tex]

Te probability is 15.6%

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