First slide

Theorems of probability

Question

A person throws a dice while he gets a number greater than 2. The probability that he gets a 6 in the last throw is

Moderate

A

$\frac{2}{3}$
B

$\frac{1}{4}$
C

$\frac{1}{3}$
D

$\frac{1}{12}$

Solution

Let E₁, be the event that six shows when a dice is thrown.
Let E₂be the event that number less than or equal to 2 shows
when a dice is thrown.
$P (E_{1}) = 1 / 6 and P (E_{2}) = 2 / 6 = 1 / 3$
E = the event that six turns up in the last throw = the event that E₁ happens in the all previous throws and E₂happens in the last throw
$\begin{array}{l} P (E) = P (E_{1} E_{2} or E_{1} E_{1} E_{2} or E_{1} E_{1} E_{1} E_{2} \dots . .) \\ = P (E_{1}) \cdot P (E_{2}) + P (E_{1}) \cdot P (E_{1}) \cdot P (E_{2}) \\ + P (E_{1}) \cdot P (E_{1}) \cdot P (E_{1}) \cdot P (E_{2}) + \dots \\ = \frac{1}{3} \cdot \frac{1}{6} + {(\frac{1}{3})}^{2} \cdot \frac{1}{6} + {(\frac{1}{3})}^{3} \cdot \frac{1}{6} + \dots \\ P (E) = \frac{1}{3} \cdot \frac{1}{6} \times \frac{1}{1 - 1 / 3} \Rightarrow P (E) = \frac{1}{12} \end{array}$

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