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Grade 12th passMagical Mathematics[Interesting Approach]

This is a probability question. Suppose we start our journey from step number 0. We now toss a coin. If the result is a head, we add 3 to a counter. If the result is a tail, we add 2 to the counter. We now move to step number 1. We toss the coin again and depending on the result, we update the counter. We then move on to step number 2 and repeat the experiment and so on. The question is, when we arrive at step 10, what is the probability that the counter reads 23?

Profile image of Nirmal
10 Years agoGrade 12th pass
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1 Answer

Profile image of Riddhish Bhalodia
10 Years ago
There are total of 11 steps from 0 to 10 so if lets suppose at every counter there was tails and hence the counter increments by 2 at every step and hence the total would be 22. It is an obvious reasoning from here that to get 23 as the counter value there must be an heads toss at ONLY ONE out of total 11 steps.

So the probablity of having 10 tails and 1 head sequenceis

P = \binom{11}{1} \frac{1}{2^{11}} = \frac{11}{2^{11}}