Question

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?

Answer 1

a) 0.076

b) 0.5

c) 0.084

d) 0.25

e) 0.00024

f) 0.156

Explanation:

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

`P=(n)/(N)`

n: options for an ace = 4

N: total number of cards = 52

`P=(4)/(52)=0.076`

The probability is 7.6%

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

`P=(50)/(100)=0.5`

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:

`P=(31)/(365)=0.084`

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.

`P=P_(odd)P_(pair)=((1)/(2))((1)/(2))=(1)/(4)=0.25`

The probability is 25%

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

`P=(1)/(2)*(1)/(2)*(1)/(2)*(1)/(2)*(1)/(2)*(1)/(2)=0.00024`

Probability of 0.024%

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

`P=((1)/(2))^5=(1)/(32)`

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

`P=5((1)/(32))=(5)/(32)=0.156`

Te probability is 15.6%