In this blog, we will solve several practice questions on probability, covering concepts like dice rolls, coin tosses, and permutations. These questions are designed to reinforce your understanding of probability principles, with detailed explanations for each solution.

Question 1: Four Dice Rolls with Even Outcomes

Problem: A fair six-sided die is rolled four times. What is the probability that all four outcomes are even numbers?

Solution:
– Even numbers on a die = {2, 4, 6} → 3 options.
– Total outcomes = 6 × 6 × 6 × 6 = 6⁴ = 1296.
– Favorable outcomes: Assume we need to form a 4-digit sequence using the numbers 2, 4, 6. Each of the four positions can be filled with 3 numbers (2, 4, or 6).
– Favorable outcomes = 3 × 3 × 3 × 3 = 3⁴ = 81.
– Probability = 81 / 1296 = 1 / 16.
– For more on permutations and combinations, refer to our course: Basics of Counting in Engineering Mathematics.

Answer: 1/16

Question 2: 3-Digit Number with Identical Digits

Problem: A 3-digit number is formed by choosing digits randomly from 1 to 9 (no zero). What is the probability that all digits are the same?

Solution:
– Total 3-digit numbers (from 100 to 999) using 1–9 only = 9 × 9 × 9 = 729.
– Numbers with all same digits: 111, 222, …, 999 → 9 such numbers.
– Probability = 9 / 729 = 1 / 81.

Answer: 1/81

Special Note on Coin Tosses

Note:
– In the case of coins, we can use the combination formula directly without needing to apply the permutation of alike items formula.
– Example: A coin is tossed 10 times. Find the number of outcomes in which exactly 4 heads appear.

Solution:
– We need 4 heads, so by default, there will be 6 tails. There are two ways to calculate the number of such outcomes:

• a) Using permutation of alike items: 10! / (4! × 6!) = 210
• b) Using combinations: $C_4^{10}$ = 210 (choosing 4 positions for heads) or $C_6^{10}$ = 210 (choosing 6 positions for tails).

– All three approaches give the same answer: 210.

Answer: 210

Question 3: Five Coin Tosses with Exactly 3 Heads

Problem: A coin is tossed 5 times. What is the probability that exactly 3 heads occur?

Solution:
– Total outcomes = 2 × 2 × 2 × 2 × 2 = 2⁵ = 32.
– The 3 heads can be at any position, e.g., {HHHTT, HTHTH, THHHT, TTHHH, …}.
– Ways to choose 3 positions for heads out of 5 positions = $C_3^5$ = 10.
– Probability = 10 / 32 = 5 / 16.

Answer: 5/16

Question 4: 3-Letter Password with No Repeated Letters

Problem: You pick a 3-letter password using only the letters A, B, and C. What is the probability that no letter is repeated?

Solution:
– Total possible passwords = 3 × 3 × 3 = 27.
– This is because each of the 3 positions can be filled by any of the 3 letters: A, B, or C.
– For passwords with no repeated letters, we are selecting and arranging 3 different letters in 3 positions:

• For the first position, we have 3 choices (A, B, or C).
• For the second position, only 2 letters remain (since one is already used in the first position).
• For the third position, only 1 letter remains (since two letters are already used).

– Number of favorable outcomes = 3 × 2 × 1 = 6.
– Probability = 6 / 27 = 2 / 9.
– Example of passwords with no repeated letters: ABC, ACB, BAC, BCA, CAB, CBA (total 6).

Answer: 2/9

Question 5: Product of Two Dice as a Perfect Square

Problem: Two dice are thrown simultaneously. Find the probability that the product of the numbers on the upper faces is a perfect square.

Solution:
– Total outcomes = 6 × 6 = 36.
– Favorable outcomes (product is a perfect square): {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6), (1,4), (4,1)}.
– Explanation: A product is a perfect square if it is 1, 4, 9, 16, 25, or 36. Checking pairs:

• 1 × 1 = 1
• 2 × 2 = 4
• 3 × 3 = 9
• 4 × 4 = 16
• 5 × 5 = 25
• 6 × 6 = 36
• 1 × 4 = 4, 4 × 1 = 4

– Total favorable outcomes = 8.
– Probability = 8 / 36 = 2 / 9.

Answer: 2/9

Question 6: Six Coin Tosses

Problem: A coin is tossed 6 times. Find:
(a) The probability that all outcomes are identical.
(b) The probability that heads and tails appear at least once.

Solution (a):
– Total outcomes = 2 × 2 × 2 × 2 × 2 × 2 = 2⁶ = 64.
– Identical outcomes = {HHHHHH, TTTTTT} = 2.
– Probability = 2 / 64 = 1 / 32.

Solution (b):
– Manual counting or applying any formula can make the question difficult to solve. When the keyword “at least” is present, we can use a shortcut:
– Favorable outcomes = Total outcomes – outcomes where heads or tails do not appear (i.e., only heads or only tails appear).
– Total outcomes = 2⁶ = 64.
– Outcomes where only heads or only tails appear = {HHHHHH, TTTTTT} = 2.
– Favorable outcomes = 64 − 2 = 62.
– Probability = 62 / 64 = 31 / 32.

Answer (a): 1/32
Answer (b): 31/32

Question 7: Four Coin Tosses with Unequal Heads and Tails

Problem: A coin is tossed 4 times. What is the probability that the number of heads is not equal to the number of tails?

Solution:
– Total outcomes = 2⁴ = 16.
– Equal heads and tails: Only when 2 heads and 2 tails → $C_2^4$ = 6.
– Outcomes where heads ≠ tails = 16 − 6 = 10.
– Probability = 10 / 16 = 5 / 8.

Answer: 5/8

Question 8: Ten Coin Tosses with Equal Heads and Tails

Problem: A coin is tossed 10 times. Find the probability that an equal number of heads and tails appear.

Solution:
– We need 5 heads and 5 tails. This means we are looking for outcomes where exactly 5 heads appear — the remaining 5 tosses will automatically be tails.
– Favorable outcomes (exactly 5 heads) = $C_5^{10}$ = 252.
– Total outcomes = 2¹⁰ = 1024.
– Probability = 252 / 1024 = 63 / 256.

Answer: 63/256

Conclusion

These practice questions cover key probability concepts, including permutations, combinations, and binomial probabilities. Practice more to master these techniques!