A box contains 6 red marbles numbers from 1 through 6 and 4 white marbles 12 through 15. Find the probability that a marble drawn 'at random' is white and odd numbered.

Question

A box contains 6 red marbles numbers from 1 through 6 and 4 white marbles 12 through 15. Find the probability that a marble drawn 'at random' is white and odd numbered. (A) 5 (B)  (1/5) (C) 6 (D)  (1 /6)

Tardigrade Admin · Accepted Answer

Total number of marbles, n(S)=6+4=10 Marbles which are white and odd numbered =2 ∴ n(E)=2 Now, P(E)=(n(E)/n(S))=(2/10) =(1/5)

Q. A box contains $6$ red marbles numbers from $1$ through $6$ and $4$ white marbles $12$ through $15$ . Find the probability that a marble drawn 'at random' is white and odd numbered.

$5$

$\frac{1}{5}$

$6$

$\frac{1}{6}$

Total number of marbles, $n (S) = 6 + 4 = 10$
Marbles which are white and odd numbered $= 2$
$∴ n (E) = 2$
Now, $P (E) = \frac{n ( E )}{n ( S )} = \frac{2}{10}$
$= \frac{1}{5}$

Q. A box contains 6 red marbles numbers from 1 through 6 and 4 white marbles 12 through 15. Find the probability that a marble drawn 'at random' is white and odd numbered.

5

51​

6

61​

Total number of marbles, n(S)=6+4=10 Marbles which are white and odd numbered =2 ∴n(E)=2 Now, P(E)=n(S)n(E)​=102​ =51​

Q. A box contains $6$ red marbles numbers from $1$ through $6$ and $4$ white marbles $12$ through $15$ . Find the probability that a marble drawn 'at random' is white and odd numbered.

$5$

$\frac{1}{5}$

$6$

$\frac{1}{6}$

Total number of marbles, $n (S) = 6 + 4 = 10$
Marbles which are white and odd numbered $= 2$
$∴ n (E) = 2$
Now, $P (E) = \frac{n ( E )}{n ( S )} = \frac{2}{10}$
$= \frac{1}{5}$