00:01
So in this problem bob is fixing two clocks, that is two alarm class to wake up in the morning, right? and the percentage of waking up is 46 by 100, that is 46 percentage, so which is equal to 0 .46.
00:16
So now the probability of waking up by the first clock is 0 .46.
00:22
And similarly, the probability of waking up by the second clock is also 0 .46.
00:27
But the question is he is not waking up for both the clocks he is sleeping okay so that is we have to find p of a union b the whole bar whole complement we have to find it out so from this p of a complement is equal to 1 minus p of a which is equal to 0 .54 and p of b complement is equal to 1 minus p of b which is equal to 0 .54 then based on this de marganclay here p of a union b, the whole complement is equal to p of a complement, intersection b complement.
01:07
Because here a and b are two different clocks, they two are disjoint.
01:12
Therefore, b of a complement, intersection b complement can be written as p of a complement multiplied by p of b complement.
01:20
So which is equal to 0 .54, the whole square, which is equal to 0 .2916.
01:27
Then the next question is, he is fixing three clocks to wake up.
01:32
Okay, so now the clocks increased three clocks to wake up.
01:35
What is the probability of getting up at least by one clock, that is p of a union, b, union c, which is equal to p of a plus it is given that it is at least 82%.
01:49
Okay, so waking up by any one of the three clock is at least 82 percentage.
01:54
So p of a plus p of b plus p of c minus p of a intersection b minus p of b intersection c minus p of a intersection c plus p of a intersection b intersection c...