Cast spells with numbers and discover the magic of chance!
Write the sample space for tossing three coins using tree diagram.
Start with the first coin toss (H or T), then branch for second toss from each outcome, then branch again for third toss.
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
Write the sample space for selecting two balls from a bag containing balls numbered 1 to 6 (using tree diagram).
We're selecting 2 balls at a time from balls numbered 1-6.
Since order doesn't matter, (1,2) is same as (2,1).
S = {(1,2), (1,3), (1,4), (1,5), (1,6), (2,3), (2,4), (2,5), (2,6), (3,4), (3,5), (3,6), (4,5), (4,6), (5,6)}
Total number of outcomes = 15
If P(A):P(Ā) = 17:15 and n(S) = 640, find (i) P(Ā) (ii) n(A).
Given P(A):P(Ā) = 17:15
Let P(A) = 17x and P(Ā) = 15x
17:15 Ratio Visualization
We know P(A) + P(Ā) = 1
So, 17x + 15x = 1 ⇒ 32x = 1 ⇒ x = 1/32
(i) P(Ā) = 15x = 15/32
(ii) P(A) = 17/32
Given n(S) = 640, so n(A) = P(A) × n(S) = (17/32) × 640 = 340
A coin is tossed thrice. What is the probability of getting two consecutive tails?
Sample space for 3 coin tosses (8 outcomes):
Outcomes with two consecutive tails:
{HTT, TTH, TTT}
(Note: TTT has two pairs of consecutive tails but we count it once)
Number of favorable outcomes = 3
Total outcomes = 8
Probability = 3/8
Cards numbered 1 to 1000 are in a box. Players select one card (not replaced). If the card has a perfect square >500, they win. Find probability that:
(i) first player wins (ii) second player wins if first won
Find numbers n where n² > 500 and n² ≤ 1000
23² = 529, 24² = 576, ..., 31² = 961 (since 32² = 1024 > 1000)
Number of perfect squares >500 = 31 - 22 = 9
9 Winning Numbers
(i) P(first player wins) = 9/1000
If first player won, one perfect square is removed, and total cards are now 999
Remaining perfect squares = 8
(ii) P(second player wins) = 8/999
A bag has 12 blue and x red balls. (i) Find P(red). (ii) If adding 8 red balls makes P(red) twice the original, find x.
Total balls = 12 (blue) + x (red) = 12 + x
(i) P(red) = x/(12 + x)
New number of red balls = x + 8
New total balls = 12 + x + 8 = 20 + x
New P(red) = (x + 8)/(20 + x)
According to problem: new P(red) = 2 × original P(red)
(x + 8)/(20 + x) = 2 × [x/(12 + x)]
(x + 8)(12 + x) = 2x(20 + x)
x² + 20x + 96 = 40x + 2x²
0 = x² + 20x - 96
x = [-20 ± √(400 + 384)]/2 = [-20 ± √784]/2 = [-20 ± 28]/2
Taking positive solution: x = 8/2 = 4
So, x = 4
Two unbiased dice are rolled once. Find probability of getting:
(i) a doublet (ii) product as prime (iii) sum as prime (iv) sum as 1
When two dice are rolled, total outcomes = 6 × 6 = 36
Two dice like these rolled together
(i) Doublet: (1,1), (2,2), ..., (6,6) → 6 outcomes
Probability = 6/36 = 1/6
(ii) Product is prime: Only possible when one die shows 1 and the other shows a prime number (2,3,5)
Possible outcomes: (1,2), (1,3), (1,5), (2,1), (3,1), (5,1) → 6 outcomes
Probability = 6/36 = 1/6
(iii) Sum as prime number: Possible sums are 2,3,5,7,11
Number of ways:
Sum 2: (1,1) → 1
Sum 3: (1,2), (2,1) → 2
Sum 5: (1,4), (2,3), (3,2), (4,1) → 4
Sum 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6
Sum 11: (5,6), (6,5) → 2
Total favorable = 1 + 2 + 4 + 6 + 2 = 15
Probability = 15/36 = 5/12
(iv) Sum as 1: Impossible (minimum sum is 2)
Probability = 0/36 = 0
Three fair coins are tossed together. Find the probability of getting:
(i) All heads (ii) At least one tail (iii) At most one head (iv) At most two tails
When tossing 3 coins, each has 2 outcomes (H or T), so total outcomes = 2 × 2 × 2 = 8
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
(i) All heads: Only {HHH} → 1 outcome
Probability = 1/8
(ii) At least one tail: All except HHH → 7 outcomes
Probability = 7/8
(iii) At most one head: {TTT, HTT, THT, TTH} → 4 outcomes
Probability = 4/8 = 1/2
(iv) At most two tails: All except TTT → 7 outcomes
Probability = 7/8
A bag contains 5 red balls, 6 white balls, 7 green balls, 8 black balls. One ball is drawn at random. Find the probability that the ball drawn is:
(i) White (ii) Black or red (iii) Not white (iv) Neither white nor black
Total = Red + White + Green + Black = 5 + 6 + 7 + 8 = 26 balls
(i) P(White) = Number of white balls / Total = 6/26 = 3/13
(ii) P(Black or Red) = (8 + 5)/26 = 13/26 = 1/2
(iii) P(Not white) = 1 - P(White) = 1 - 6/26 = 20/26 = 10/13
(iv) P(Neither white nor black) = P(Red or Green) = (5 + 7)/26 = 12/26 = 6/13
In a box there are 20 non-defective and some defective bulbs. If the probability of selecting a defective bulb is 3/8, find the number of defective bulbs.
Let number of defective bulbs = x
Total bulbs = Non-defective + Defective = 20 + x
P(Defective) = Number of defective / Total bulbs
3/8 = x / (20 + x)
Cross multiply: 3(20 + x) = 8x
60 + 3x = 8x
60 = 5x
x = 12
There are 12 defective bulbs in the box.
Some boys are playing a game where a stone landing in a circular region is a win. What is the probability to win the game? (π = 3.14)
This problem requires knowing the area of the circular target region and the total possible landing area.
Probability = (Area of circular target) / (Total possible landing area)
For example, if the target has radius r and the stone can land anywhere in a square of side 2r:
Area of circle = πr²
Area of square = (2r)² = 4r²
Probability = πr²/4r² = π/4 ≈ 3.14/4 = 0.785 or 78.5%
Two customers visit a shop on any day from Monday to Saturday. What is the probability that both will visit on:
(i) The same day (ii) Different days (iii) Consecutive days
Each customer has 6 choices (Mon-Sat)
Total outcomes = 6 × 6 = 36
Each customer can independently choose any of these 6 days.
(i) Same day: (Mon,Mon), (Tue,Tue), etc. → 6 outcomes
Probability = 6/36 = 1/6
(ii) Different days: Total - Same day = 36 - 6 = 30
Probability = 30/36 = 5/6
(iii) Consecutive days:
(Mon,Tue), (Tue,Mon), (Tue,Wed), (Wed,Tue), (Wed,Thu), (Thu,Wed), (Thu,Fri), (Fri,Thu), (Fri,Sat), (Sat,Fri)
Total = 10 outcomes
Probability = 10/36 = 5/18
In a game with ₹150 entry fee, Dhana tosses a coin 3 times. If she gets:
• 1-2 heads: gets entry fee back
• 3 heads: double entry fee
• Otherwise: loses
Find the probability that she (i) gets double fee (ii) gets entry back (iii) loses.
Sample space for 3 coin tosses (8 outcomes):
(i) Gets double fee (3 heads): {HHH} → 1 outcome
Probability = 1/8
(ii) Gets entry back (1 or 2 heads):
1 head: {HTT, THT, TTH} → 3 outcomes
2 heads: {HHT, HTH, THH} → 3 outcomes
Total = 6 outcomes
Probability = 6/8 = 3/4
(iii) Loses entry fee (0 heads): {TTT} → 1 outcome
Probability = 1/8