Find the probability of selecting none of the correct six integers in a lottery, where the order in which these integers are selected does not matter, from the positive integers not exceeding the given integers. (Enter the value of probability in decimals. Round the answer to two decimal places.) 40

Probability of selecting none of the correct six integers:

a) 0.350

b) 0.427

c) 0.489

d) 0.540

Step-by-step explanation:

a) 40

Given:

Number of integers in a lottery  6

Order in which these integers are selected does not matter

To find:

Probability of selecting none of the correct six integers

Solution:

When the order of selection does not matter then we use Combinations.

Given integers = 40

Number of ways to choose 6 from 40.

Let A be the sample space of choosing digits 6 from 40.

Then using Combinations:

(n,k) = n! / r! (n-r)!

n = 40

r = 6

40C6

=(40,6) = 40! / 6! ( 40 - 6)!

           = 40! / 6!34!

           = 40*39*38*37*36*35*34! / 6!34!

           = 2763633600 / 720

          = 3838380

Let E be the event of selecting none of the correct six integers.

So using combinations we can find the total number of ways of selecting none of 6 integers from 40

n = 40 - 6 = 34

r = 6

34C6

=(34,6) = 34! / 6! ( 34 - 6)!

           = 34! / 6! 28!

           = 34 * 33 * 32 * 31 * 30 * 29 * 28! / 6! 28!

           =968330880 / 720

           = 1344904

Probability of selecting none of the correct six integers:

P(E) = E / A

       = 1344904 / 3838380

        = 0.350

Probability of selecting none of the correct six integers is 0.350

b) 48

Following the method used in part a)

(n,k) = n! / r! (n-r)!

n = 48

r = 6

48C6

=(48,6) = 48! / 6! ( 48 - 6)!

            = 48! / 6! ( 42 )!

            = 48*47*46*45*44*43*42! / 6!42!

            = 8835488640 / 720

            = 12271512

Let E be the event of selecting none of the correct six integers.

So using combinations we can find the total number of ways of selecting none of 6 integers from 48

n = 48 - 6 = 42

r = 6

42C6

= (42,6) = 42! / 6! ( 42 - 6)!

              = 42! / 6! 36!

              = 3776965920

              = 5245786

P(E) = E / A

       = 5245786/12271512

       = 0.427

c) 56

(n,k) = n! / r! (n-r)!

n = 56

r = 6

56C6

=(56,6) = 56! / 6! ( 56- 6)!

            = 56! / 6! ( 50 )!

            = 56*55*54*53*52*51*50! / 6! 50!

            = 23377273920/6

            = 32468436

Let E be the event of selecting none of the correct six integers.

So using combinations we can find the total number of ways of selecting none of 6 integers from 56

n = 56 - 6 = 50

(50,6) = 50! / 6! ( 50- 6)!

           = 50*49*48*47*46*45*44! / 44! 6!

           = 11441304000 / 6

           = 15890700

P(E) = E / A

       = 15890700 / 32468436

      = 0.489

d) 64

(n,k) = n! / r! (n-r)!

n = 64

r = 6

64C6

=(64,6) = 64! / 6! ( 64 - 6)!

            = 64! / 6! ( 58 )!

            = 64*63*62*61*60*59*58! / 6! 58!

            = 53981544960 / 720

            = 74974368

Let E be the event of selecting none of the correct six integers.

So using combinations we can find the total number of ways of selecting none of 6 integers from 64

n = 64 - 6 = 58

(58,6) = 58! / 6! ( 58- 6)!

           = 58*57*56*55*54*53*52! / 52! 6!

           = 29142257760/ 6

           = 40475358

P(E) = E / A

       = 40475358/ 74974368

      = 0.540

bull

step-by-step explanation:

yes the total will be same because in both of them the expression.is rearranged but their sign remains the same hence it hss just been shuffled