We have a deck of 10 cards numbered from 1-10. Some are grey and some are white. The cards numbered are 1,2,3,5,6,8 and 9 are grey. The cards numbered 4,7, and 10 are white. A card is drawn at random. Let X be the event that the drawn card is grey, and let P(X) be the probability of X. Let not X be the event that the drawn card is not grey, and let P(not X) be the probability of not X.

We have a deck of 10 cards numbered from 110 Some are grey and some are white The cards numbered are 123568 and 9 are grey The cards numbered 47 and 10 are whit class=

Respuesta :

Given:

The cards numbered are, 1,2,3,5,6,8, and 9 are grey.

The cards numbered 4,7 and 10 are white.

The total number of cards =10.

Let X be the event that the drawn card is grey.

P(X) be the probability of X.

Required:

We need to find P(X) and P(not X).

Explanation:

All possible outcomes = All cards.

[tex]n(S)=10[/tex]

Click boxes that are numbered 1,2,3,5,6,8, and 9 for event X.

The favourable outcomes = 1,2,3,5,6,8, and 9

[tex]n(X)=7[/tex]

Since X be the event that the drawn card is grey.

The probability of X is

[tex]P(X)=\frac{n(X)}{n(S)}=\frac{7}{10}[/tex]

Let not X be the event that the drawn card is not grey,

All possible outcomes = All cards.

[tex]n(S)=10[/tex]

Click boxes that are numbered 4,7, and 10 for event not X.

The favourable outcomes = 4,7, and 10

[tex]n(not\text{ }X)=3[/tex]

Since not X be the event that the drawn card is whic is not grey.

The probability of not X is

[tex]P(not\text{ }X)=\frac{n(not\text{ }X)}{n(S)}=\frac{3}{10}[/tex]

Consider the equation.

[tex]1-P(not\text{ X\rparen}[/tex][tex]Substitute\text{ }P(not\text{ }X)=\frac{3}{10}\text{ in the equation.}[/tex][tex]1-P(not\text{ X\rparen=1-}\frac{3}{10}[/tex][tex]1-P(not\text{ X\rparen=1}\times\frac{10}{10}\text{-}\frac{3}{10}=\frac{10-3}{10}=\frac{7}{10}[/tex]

[tex]1-P(not\text{ X\rparen is same as }P(X).[/tex]

Final answer:

[tex]1-P(not\text{ X\rparen is same as }P(X).[/tex]

Ver imagen AyriannaA106938