Question

Ram is tossing a coin. If head comes then 10 points will be given and if tail comes then 5 points will be given. If the probability of getting exactly 30 point is \( \dfrac{m}{n} \), then \( (m+n) \) equals (Where \( m \) \& \( n \) are co-prime numbers).

Hint:

When score-based coin toss questions are given, first form a linear equation using the point values, then list all valid integer solutions and add the probabilities of each case.

Answer 1

Correct Answer: 107

Step 1: Form the equation for total points.
Suppose Ram tosses the coin \( n \) times. Let the number of heads be \( h \) and the number of tails be \( t \).
Then,
\[ h+t=n \] Since each head gives \( 10 \) points and each tail gives \( 5 \) points, total score is
\[ 10h+5t=30 \] Dividing by \( 5 \), we get
\[ 2h+t=6 \]
Step 2: Find all possible non-negative integral solutions.
We need all non-negative integer solutions of
\[ 2h+t=6 \] Possible values are:
If \( h=0 \), then \( t=6 \)
If \( h=1 \), then \( t=4 \)
If \( h=2 \), then \( t=2 \)
If \( h=3 \), then \( t=0 \)
So the possible cases are:
\[ (h,t)=(0,6),(1,4),(2,2),(3,0) \]
Step 3: Calculate probability for each case.
Now we calculate the probability of each case separately.
Case 1: \( (h,t)=(0,6) \)
This means all 6 tosses are tails. Probability is
\[ \left(\frac{1}{2}\right)^6=\frac{1}{64} \] Case 2: \( (h,t)=(1,4) \)
Total tosses \( =5 \). Number of ways to choose 1 head among 5 tosses is
\[ {}^5C_1=5 \] So probability is
\[ 5\left(\frac{1}{2}\right)^5=\frac{5}{32} \] Case 3: \( (h,t)=(2,2) \)
Total tosses \( =4 \). Number of ways to choose 2 heads among 4 tosses is
\[ {}^4C_2=6 \] So probability is
\[ 6\left(\frac{1}{2}\right)^4=\frac{6}{16}=\frac{3}{8} \] Case 4: \( (h,t)=(3,0) \)
This means all 3 tosses are heads. Probability is
\[ \left(\frac{1}{2}\right)^3=\frac{1}{8} \]
Step 4: Add all probabilities.
Therefore, required probability is
\[ \frac{1}{64}+\frac{5}{32}+\frac{3}{8}+\frac{1}{8} \] Taking LCM \( =64 \), we get
\[ \frac{1}{64}+\frac{10}{64}+\frac{24}{64}+\frac{8}{64} =\frac{43}{64} \] Thus,
\[ \frac{m}{n}=\frac{43}{64} \] So,
\[ m+n=43+64=107 \]