195 views
For which of the following events will the number of outcomes exceed $50?$(Indicate all such events.)
1. The number of outcomes in which at least three heads appear in $6$ consecutive tosses of a fair coin.
2. The number of outcomes in which the sum of the digits that appear on the facing side is odd when a fair die rolled thrice.
3. The number of ways of posting $6$ different letters in $2$ different post boxes such that at least one letter is posted in each of the boxes.
4. The number of outcomes in which all the vowels appear together when the letters of the word 'PRIORITY' are reordered.

## 1 Answer

Option A:
The number of outcomes in which at least three heads appear in $6$ consecutive tosses of a fair coin.
A coin tossed $6$ times consecutively could result in outcomes ranging from $0$ heads to $6$ heads. We have to count the outcomes in which at least three heads appear;
What does at least three heads mean?
A minimum of $3$ heads. i.e., the outcomes in which we get $3$ heads or $4$ heads or $5$ heads or all $6$ as heads.

• Number of outcomes with $3$ heads: $6$ choose $3$ or $6 \mathrm{C} 3=6 ! / 3 ! \times 3 !=20$.
• Number of outcomes with $4$ heads: $6$ choose $4$ or $6 \mathrm{C} 4=6 ! / 4 ! \times 2 !=15$.
• Number of outcomes with $5$ heads: $6$ choose $5$ or $6 C 5=6 ! / 5 ! \times 1 !=6$.
• Number of outcomes with $6$ heads: $6$ choose $6$ or $6 C 6=1$ way.

So, the number of outcomes in which at least $3$ heads will appear $=20+15$ $+6+1=42$

Option A is NOT one of the answers.

Option B:
The number of outcomes in which the sum of the digits that appear on the facing side is odd when a fair die is rolled thrice.

We can either have all three times Odd numbers, OR we can have exactly one Odd number.

• When we have all three times Odd numbers: $3 \times 3 \times 3=27$ Ways.
• When we have exactly one Odd number: $(3 \mathrm{C} 1) \times 3 \times 3 \times 3=81$

Final answer $=81+27=108$

Choice $B$ is one of the answers where the number of outcomes is more than $50.$

Option C:
The number of ways of posting $6$ different letters in $2$ different post boxes such that at least one letter is posted in each of the boxes.
Let us say the letters are addressed to $A, B, C, D, E$, and $F$, and the boxes are $1$ and $2 .$

Letter $A$ can be posted in the boxes in $2$ ways. Either into box $1$ or into box $2.$

Similarly, the letter $B$ and the remaining letters can be posted into a box in $2$ ways each.

Letter $A$ can be posted in $2$ ways AND letter $B$ can be posted in $2$ ways and so on.

So, the $6$ letters can be posted in $2 \times 2 \times 2 \times 2 \times 2 \times 2=26=64$ ways.

The only problem is that these $64$ ways include the possibility that all the $6$ letters are posted into only one of the post boxes - either all $6$ into box $1$ or all $6$ into box $2 .$

If we count these two possibilities out, then we have $64-2=62$ ways in which $6$ different letters can be posted into $2$ different post boxes such that at least one letter is posted in each of the boxes.
Option $\mathrm{C}$ is one of the choices in which the number of outcomes exceeds $50.$

Option D:
The number of outcomes in which the vowels appear together when the letters of the word 'PRIORITY' are reordered.
PRIORITY is an $8$-letter word.

• The vowels in the word are $\mathrm{I}, \mathrm{O}$, and $\mathrm{I}$.
• The consonants in the word are $\mathrm{P}, \mathrm{R}, \mathrm{R}$, $\mathrm{T}$, and $\mathrm{Y}$.
• The condition to be satisfied is that the vowels appear together.

So, if we consider the $3$ vowels as one unit, the vowels will appear together. Let us call this unit comprising $3$ vowels as $\delta$.

So, we need to reorder $\mathrm{P}, \mathrm{R}, \mathrm{R}, \mathrm{T}, \mathrm{Y}$ and $\delta$
i.e., $6$ letters have to be reordered - out of which two letters are 'R's.

So, these can be reordered in $6 ! / 2 !=360$ ways.

Now the unit $\delta$ comprising $3$ letters I, $O$, and I can also be reordered.

These $3$ letters of which two are alike, can be reordered in $3 ! / 2 !=3$
ways.

So, the total number of ways to reorder the letters of the word
PRIORITY such that the vowels appear together $=360 * 3=1080$ ways.

Option D is one of the choices in which the number of outcomes exceeds $50.$

Answer:

5 votes
3 answers
1
420 views
3 votes
1 answer
2
301 views
8 votes
1 answer
3
5 votes
1 answer
4