Question

Let \(X\) denote the number of heads in a simultaneous toss of three coins, then \[ P(0<X<3) \] is:

• A. \(\dfrac12\)
• B. \(\dfrac34\)
• C. \(\dfrac78\)
• D. \(1\)

Hint:

For three coin tosses: \[ \text{Total outcomes}=2^3=8 \]

Answer 1

The Correct Option is B

Step 1: Find total outcomes When three coins are tossed simultaneously: \[ \text{Total outcomes}=2^3=8 \] Sample space: \[ \{HHH,\ HHT,\ HTH,\ THH,\ HTT,\ THT,\ TTH,\ TTT\} \]
Step 2: Understand the condition Given: \[ 0<X<3 \] where \(X\) is the number of heads. Thus: \[ X=1 \quad \text{or} \quad X=2 \]
Step 3: Count favourable outcomes For: \[ X=1 \] outcomes are: \[ HTT,\ THT,\ TTH \] Number of outcomes: \[ 3 \] For: \[ X=2 \] outcomes are: \[ HHT,\ HTH,\ THH \] Number of outcomes: \[ 3 \] Hence total favourable outcomes: \[ 3+3=6 \]
Step 4: Find probability \[ P(0<X<3) = \frac{6}{8} \] \[ = \frac34 \] Option analysis:
• Option (A): Incorrect
• Option (B): Correct
• Option (C): Incorrect
• Option (D): Incorrect Therefore: \[ \boxed{\text{(B)}} \]