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Question:You are given a function foo() that represents a biased coin. When foo() is called, it returns 0 with 60% probability, and 1 with 40% probability. Write a new function that returns 0 and 1 with 50% probability each. Your function should use only foo(), no other library method.

We know foo() returns 0 with 60% probability. How can we ensure that 0 and 1 are returned with 50% probability?

If we can somehow get two cases with equal probability, then we are done. We call foo() two times. Both calls will return 0 with 60% probability. So the two pairs (0, 1) and (1, 0) will be generated with equal probability from two calls of foo(). Let us see how.

**(0, 1): The probability to get 0 followed by 1 from two calls of foo() = 0.6 * 0.4 = 0.24
(1, 0): The probability to get 1 followed by 0 from two calls of foo() = 0.4 * 0.6 = 0.24**

So the two cases appear with equal probability. The idea is to return consider only the above two cases, return 0 in one case, return 1 in other case. For other cases [(0, 0) and (1, 1)], recur until you end up in any of the above two cases.

The below program depicts how we can use foo() to return 0 and 1 with equal probability.

#include <stdio.h> int foo() // given method that returns 0 with 60% probability and 1 with 40% { // some code here } // returns both 0 and 1 with 50% probability int my_fun() { int val1 = foo(); int val2 = foo(); if (val1 == 0 && val2 == 1) return 0; // Will reach here with 0.24 probability if (val1 == 1 && val2 == 0) return 1; // // Will reach here with 0.24 probability return my_fun(); // will reach here with (1 - 0.24 - 0.24) probability } int main(void) { printf ("%d ", my_fun()); return 0; }