Question
A game consists of tossing three coins once and then
rolling two dice. Find the probability of getting no heads in the coin toss and a sum equal to 4 on the dice.Solution
For tossing 3 coins: Total number of possible outcomes = 23 = 8 Number of favourable outcomes = 1 (TTT) So, probability of getting no heads while tossing three coins = (1/8) For rolling 2 dice: Total number of possible outcomes = 6 × 6 = 36 Number of favourable outcomes = 3 [(1,3), (2,2), (3,1)] So, Probability of getting a sum equal to 4 while rolling 2 dice = (3/36) = (1/12) Required probability = (1/8) × (1/12) = 1/96
√10201 × √3969 - (52)² = √? + (60)²
...20% of 550 × 25% of 80 = ?2 × 22
150% of 850 ÷ 25 – 25 = ?% of (39312 ÷ 1512)
1024 ÷ 32 = 2(1/2)×?
What will come in the place of question mark (?) in the given expression?
(28/14) of 56 + ? = 920 ÷ 5
?% of 320 - 69 = 123
√? = 32% of 900 + 48% of 50
?2 × 25 = 42 × 21 + 172
18 * 9 + 25% of 120 + 50% of 200 = ?
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