When two coins are tossed simultaneously, all possible
outcomes are HH, HT, TH and TT.
Total number of possible outcomes = 4
Let E be the event of getting at least one head.
Then, the favourable outcomes are HT, TH and HH
Number of favourable outcomes = 3
∴ P (getting at least 1 head) = `"Number of favourable outcomes"/"Total number of possible outcomes "= 3/4`
Page 2
In a single throw of a die, the possible outcomes are 1, 2, 3, 4, 5 and 6.
Total number of possible outcomes = 6
Let E be the event of getting an even number.
Then, the favourable outcomes are 2, 4 and 6.
Number of favourable outcomes = 3
∴ Probability of getting an even number = P (E) = `3/6 = 1/2`
Page 3
In a single throw of a die, the possible outcomes are 1, 2, 3, 4, 5 and 6.
Total number of possible outcomes = 6
Let E be the event of getting a number less than 5.
Then, the favourable outcomes are 1, 2, 3, 4
Number of favourable outcomes = 4
∴ Probability of getting a number less than 5 = P (E) = `"Number of favourable outcome"/"Total number of possible outcomes"=4/6 = 2/3`
Page 4
In a single throw of a die, the possible outcomes are 1, 2, 3, 4, 5 and 6.
Total number of possible outcomes = 6
Let E be the event of getting a number greater than 2.
Then, the favourable outcomes are 3, 4, 5 and 6.
Number of favourable outcomes = 4
∴ Probability of getting a number greater than 2 = P (E) =`"Number of favourable outcomes"/"Total number of possible outcomes" = 4/6 = 2/3`
Page 5
In a single throw of a die, the possible outcomes are 1, 2, 3, 4, 5 and 6.
Total number of possible outcomes = 6
Let E be the event of getting a number between 3 and 6.
Then, the favourable outcomes are 4, 5
Number of favourable outcomes = 2
∴ Probability of getting a number between 3 and 6 = P (E)= `"Number of favourable outcomes "/"Total number of possible outcomes" = 2/6 = 1/3`
Page 6
A die is thrown once. Find the probability of getting a number other than 3.
Let E be the event of getting a number other than 3.
Then, the favourable outcomes are 1, 2, 4, 5 and 6.
Number of favourable outcomes = 5
∴ Probability of getting a number other than 3 = P (E) = `"Number of favourable outcomes"/"Total number of possible outcomes" = 5/6`
Concept: Concept Or Properties of Probability
Is there an error in this question or solution?
Page 7
In a single throw of a die, the possible outcomes are 1, 2, 3, 4, 5 and 6.
Total number of possible outcomes = 6
Let E be the event of getting the number 5.
Then, the favourable outcome is 5.
Number of favourable outcomes = 1
∴ Probability of getting the number 5 = P (E) = `"Number of favourable outcome"/"Total number of possible outcomes" =1/6`
Text Solution
`1/4``1/2``3/4``3/8`
Answer : C
Solution : When two coins are tossed simultaneously, all possible outcomes are HH, HT, TH, TT. <br> Total number of possible outcomes = 4. <br> Let E be the event of getting at least one head. <br> Then, E is the event of getting 1 head or 2 heads. <br> So, the favourable outcomes are HT, TH, HH. <br> Number of favourable outcomes = 3. <br> `:. ` P(getting at least one head ) ` =P(E) = 3/4`.