Mrs.Sanders' grandchildren error

Probability Level pending

Mrs. Sanders has three grandchildren, who call her regularly. One calls her every three days, one calls her every four days, and one calls her every five days. All three called her on December 31, 2016. On how many days during the next year did she not receive a phone call from any of her grandchildren?

152 146 144 80 78

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2 solutions

Parth Sankhe
Dec 13, 2018

We shall solve this using the principle of exclusion and inclusion.

n(A or B or C) = n(A) + n(B) + n(C) - [n(A&B) + n(B&C) + n(C&A)] + n(A&B&C)

Hence, the number of days they would have called = 365 3 + 365 4 + 365 5 ( 365 3 × 4 + 365 4 × 5 + 365 5 × 3 ) + 365 3 × 4 × 5 \frac {365}{3} + \frac {365}{4}+\frac {365}{5} - (\frac {365}{3×4} +\frac {365}{4×5}+\frac {365}{5×3}) + \frac {365}{3×4×5}

Note that every term will have to be an integer, hence each division will be approximated to the nearest lesser integer.

The above value comes out to be 219 219 , hence the number of no-call days = 365 219 = 146 365-219=146

Jordan Cahn
Dec 13, 2018

Relevant wiki: Principle of Inclusion and Exclusion - Multiple Sets

We use inclusion-exclusion to count the number of days on which Mrs. Sanders receives at least one call: 365 3 + 365 4 + 365 5 365 3 4 365 3 5 365 4 5 + 365 3 4 5 = 121 + 91 + 73 30 24 18 + 6 = 219 \left\lfloor\frac{365}{3} \right\rfloor + \left\lfloor\frac{365}{4} \right\rfloor + \left\lfloor\frac{365}{5} \right\rfloor - \left\lfloor\frac{365}{3\cdot 4} \right\rfloor - \left\lfloor\frac{365}{3\cdot 5} \right\rfloor - \left\lfloor\frac{365}{4\cdot 5} \right\rfloor + \left\lfloor\frac{365}{3\cdot 4\cdot 5} \right\rfloor = 121 + 91 + 73 - 30 - 24 - 18 + 6 = 219 Thus, the number of days on which she does not receive a call is 365 219 = 146 365-219 = \boxed{146} .

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