A die is thrown again and again until three 5’s are obtained. The probability of obtaining the third 5 in the seventh throw of the die is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{3125}{93312} \)
The third 5 comes on the seventh throw when exactly two of the first six throws are 5s and the seventh throw is a 5:
\( \displaystyle P={}^{6}C_2\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^4\times\frac{1}{6} \)
\( \displaystyle =\frac{15\times 625}{6^7} \)
\( \displaystyle =\frac{9375}{279936} \)
\( \displaystyle =\frac{3125}{93312} \)