# Probability – 5 ## 12 thoughts on “Probability – 5”

1. amit ujjwal |

sir I couldnt understand the 3rd example.

• catcracker |

Er, there are only two examples here?

regards
J

2. Deepak Gupta |

In the last example, initially you have considered 3 as number of days in a weekend and later 2.

• catcracker |

No, I haven’t actually…3 is the number of ways of choosing 1 friend (out of 3) to have his/her birthday on a weekend…

regards
J

3. seema |

sir, in second question, what if the question asks for exactly 3 birthdays on weekend?
What will be the difference here in exactly 3, atleast 1…?

4. seema |

because if we take three cases of atleast 1, i.e. either exactly one, or exactly 2, or exactly three brthdays on wekeedns so, 2/7+ 4/49+ 8/343, this gives 134/343. what is wrong in this approach?

5. catcracker |

2/7 is not exactly 1. 2/7 is for one particular person and it includes cases where the others may or may not have a bday on weekend. You’d have to use a Venn diagram logic (inclusion-exclusion principle) to solve it; much more painful!

regards
J

6. Shreya |

So is the answer for 2nd question: 7*(218/343)?

7. shreya24 |

• catcracker |

No, it is 218/343 only as mentioned in the original post. (By the way, notice that your answer is greater than 1, and hence can be ruled out right away…that can be a useful check in probability, remembering that the answer will always lie between 0 and 1)

regards
J

8. Adorini |

Sir,
If we take the solution for at least one birthday on a weekend as exactly one+ exactly two+ exactly three then probability will be 78/343 ( As 50/343 for exactly one, 20/343 for exactly two and 8/343 for all three).
Please explain where am i going wrong.

Regards,
• catcracker |