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14.Probability
normal
A die is loaded in such a way that each odd number is twice as likely to occur as each even number. If $E$ is the event that a number greater than or equal to $4$ occurs on a single toss of the die then $P(E)$ is equal to
A
$\frac{4}{9}$
B
$\frac{2}{3}$
C
$\frac{1}{3}$
D
$\frac{1}{2}$
Solution
Prob. of even $n \circ=p$
Prob. of odd no $=2 p$
$3(2 p)+3(p)=1 \Rightarrow p=\frac{1}{9}$
Req. prob $=\frac{4}{9}$
Standard 11
Mathematics
Similar Questions
Fill in the blanks in following table :
| $P(A)$ | $P(B)$ | $P(A \cap B)$ | $P (A \cup B)$ |
| $\frac {1}{3}$ | $\frac {1}{5}$ | $\frac {1}{15}$ | …….. |
easy
