Probability
Binomial Probability
IIT-JEE 2004
Grade 12
Question:
A man takes a step forward with probability 0.4 and backward with probability 0.6. The probability that at the end of eleven steps he is just one step away from the starting point is:
(1) $^{11}C_6 (0.4)^6 (0.6)^5$
(2) $^{11}C_6 (0.4)^6 (0.6)^5 + ^{11}C_5 (0.4)^5 (0.6)^6$
(3) $^{11}C_5 (0.4)^6 (0.6)^5$
(4) $2 \cdot ^{11}C_6 (0.4)^5 (0.6)^6$
Step-by-Step Solution
Key Concept: Being "one step away" after 11 steps can happen in two distinct ways — either 6 forward & 5 backward steps, or 5 forward & 6 backward steps — so the required probability is the sum of these two binomial probabilities.
The detailed step-by-step mathematical proof is available inside the Mathbee app workspace.
Correct Answer: (2)