# Difficult Probability Solved QuestionAptitude Discussion

 Q. Dexter was born between October $6^{th}$ and $10^{th}$ ($6^{th}$ and $10^{th}$ excluding). His year of birth is also known. What is the probability of Dexter being born on a Saturday?
 ✔ A. 0 or 3/7 ✖ B. 1/7 or 3/7 ✖ C. 1/3 or 1/7 ✖ D. Cannot be determined

Solution:
Option(A) is correct

Since the year of birth is known, the birthday being on Saturday can have a zero probability.

Also since between $6^{th}$ and $10^{th}$, there are three days
i.e. $7^{th}$,$8^{th}$ and $9^{th}$

As pointed out by Ankit in the comments, we have following possibilities on these three dates,

1. Monday, Tuesday, Wednesday
2. Tuesday, Wednesday, Thursday
3. Wednesday, Thursday, Friday
4. Thursday, Friday, Saturday,
5. Friday, Saturday, Sunday
6. Saturday, Sunday, Monday
7. Sunday, Monday, Tuesday

Out of these 7 events, we have 3 chances of his birthday falling on saturday.

$\Rightarrow \text{Probability}= \dfrac{Favorable Events}{Total Events}$

$\Rightarrow \text{Probability}= \dfrac{3}{7}$

Therefore, the probability of birthday falling on Saturday can be 3/7.

Edit: Thank you Ankit for the input.

## (8) Comment(s)

M Droy
()

If we know the year then there are only 2 possibilities - yes or no.

And the answer is 0 or 1.

If you have any value other than 0 or 1, it means you don't know the year (therefore 3/7 only)

Aastha
()

Chandrakant
()

I think the answer should be 1/7

There are three cases:

i) He was born on 7th and it was Saturday: $\left(\dfrac{1}{3}\right) \times \left(\dfrac{1}{7}\right)$

ii) He was born on 8th and it was Saturday: $\left(\dfrac{1}{3}\right) \times \left(\dfrac{1}{7}\right)$

iii) He was born on 9th and it was Saturday: $\left(\dfrac{1}{3}\right) \times \left(\dfrac{1}{7}\right)$

In total: His chances of being born on Saturday:

$3 \times$\left(\dfrac{1}{3}\right) \times \left(\dfrac{1}{7}\right)= \dfrac{1}{7}$Deepti () I guess while saying, 'He was born on 7th and it was Saturday: (1/3)*(1/7)', you are multiplying two probabilities,$P(\text{Saterday})=\dfrac{1}{7}$and$P(\text{it's 7th})=\dfrac{1}{3}$. How come$P(\text{it's 7th})$is$=\dfrac{1}{3}$. Surely you must have considered only 3 dates, i.e. 7, 8 and 9. But during any year having 7 as a date will not have$\dfrac{1}{7}$probability. I guess all the 3 dates (i.e. 7, 8 and 9) are to be looked together. Ankit () As there are 3 dates i.e. 7th, 8th and 9th we can have following probabilities 1. Monday, Tuesday, Wednesday 2. Tuesday, Wednesday, Thursday 3. Wednesday, Thursday, Friday 4. Thursday, Friday, Saturday, 5. Friday, Saturday, Sunday 6. Saturday, Sunday, Monday 7. Sunday, Monday, Tuesday Out of these 7 events, we have 3 chances of his birthday falling on saturday Therefore, probability of birthday falling on Saturday can be$\frac{3}{7}\$

Deepak
()

Thank you Ankit for pointing out, updated the solution.

Bibek Sahu
()

if date of birth is known including year,the day of his/her birth can be found out.so where come probability,its either 0 or 1.

Akash
()

it is zero and 3/7 because three consecutive days has to be selected if he is really born on saturday.

So,according to me the option itself is wrong