Question
Tom and Jerry throw alternatively a pair of dice. Tom
wins if he throws 8 before Jerry throws 7 and Jerry wins if he throws 7 before Tom throws 8. Find their respective chance of winning, if Tom begins.Solution
8 can be thrown with a pair of dice in the following ways (2,6), (6,2), (4,4), (5,3), (3,5) So, Probability of throwing ‘8’ = 5/36 And Probability of not throwing ‘8’ = 31/36 And 7 can be thrown with a pair of dice in the following ways (1,6), (6,1), (2,5), (5,2), (3,4), (4,3) So, the probability of throwing a ‘7’ = 6/36= 1/6 And Probability of not throwing ‘7’ = 5/6 Let E1 be the event of the throwing a ‘8’ in a single throw of a pair of dice and E2 be the event of throwing a ‘7’ in a single throw of a pair of Dice Then P(E1) = 5/36 , P(E2) = 1/6 And P(¯E1)= 31/36, P(¯E2) = 5/6 Tom wins if he throws ‘8’ in first or third or fifth............ throws. Probability of Tom throwing a ‘8’ in First throw = P(E1) = 5/36 And Probability of Tom throwing a ‘8’ in third throw = P(¯E1 ∩ ¯E2 ∩ E1 ) = P(¯E1) × P(¯E2) × P(E1) = 31/36 × 5/6 ×5/36 Similarly, Probability of Tom throwing a ‘8’ in fifth throw = P(¯E1) × P(¯E2) × P(¯E1) × P(¯E2) × P(E1) = (31/36)^2×(5/6)² ×5/36 Hence, Probability of winning of Tom = P[E1∪(¯E1 ∩ ¯E2 ∩ E1)∪ (¯E1∩ ¯E2 ∩ ¯E1∩¯E2∩ E1)∪……. ] = P[E1+(¯E1 ∩ ¯E2 ∩ E1)+ (¯E1∩ ¯E2 ∩ ¯E1∩¯E2∩ E1)+⋯…. ] Sum of n terms of Geometric Progression , if r < 1 = a/(1-r) ∴ Probability of winning of Tom = (5/36)/(1-(31/36 × 5/6)) = 30/61 Thus, Probability of winning of Jerry = 1 - 30/61 = 31/61
342 – 232 + 256 – 104 = ?
14 × 11 + 25 – ? = 21% of 300
2850 ÷ 2.5 - ? × 42 = 300
What will come in the place of question mark (?) in the given expression?
(72 × 6 ÷ 12) × 6 = ?
Find the value of 325 × 5 + 31 × 21 – 22 × (63 − 52).
√? = 80% of 720 - 22% of 2500
5555 ÷ 11 ÷ 5 = 100 + ?
- What will come in place of (?) in the given expression.
[45 + (36 ÷ 6)] × 2 – 10 = ? (23.95)2 – (25.006)2 + (8.0099)2 – (7.07)2 = ? - (14.990)2
If a³ - b³ = (a - b)(a² + ab + b²), find a³ - b³ when a = 10 and b = 4.