Simplify the expression: `(ab + bc)^2– 2ab²c`

Explanation

The man can be one stap away from the statting point if (i) either he is one stgep aheld or (ii) one step behind the starting point. Now if at the end of the 11 steps, the man is one step ahead of the starting point, then our of 11 steps, he must have taken six forward steps and five backward steps. The probability of this event is
`""^(11)C_(6)xx(0.4)^(6)xx(0.4)^(5)=462xx(0.6)^(6)xx(0.6)^(5)`
Again if at the end of the 11 steps, then man is one step behind the staring point, then out of 11 steps, he must have taken six backward steps and five forward steps, The probability of this event is
`""^(11)C_(6)xx(0.6)^(6)xx(0.4)^(5)=462xx(0.6)^(6)xx(0.4)^(5)`
Since the events (i) and (ii) are mutually exclusive, the required probability that one of these events hapen is
`[462xx(0.4)^(6)xx(0.6)^(5)]+[462xx(0.6)^(6)xx(0.4)^(5)]`
`{:(=462xx(0.4)^(5)xx(04)^(5)(0.4+0.6)),(=462xx(0.4xx0.6)^(5)),(=462(0.24)^(5)):}`