In the previous section we completed the discussion on the probability related to drawing a coloured ball, tossing a coin, rolling a die, drawing a card etc., We saw some solved examples also. In this section, we will see some solved examples with a higher level of difficulty.
Solved example 1.10
Tickets marked with numbers from 1 to 25 are put in a box and mixed well. Then one ticket is drawn from the box without looking. What is the probability that the drawn ticket is a multiple of 3 or 7?
Solution:
Step 1: Write the outcomes:
Outcome 1: The drawn ticket is the one which is marked with '1'
Outcome 2: The drawn ticket is the one which is marked with '2'
Outcome 3: The drawn ticket is the one which is marked with '3'
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- - -
Outcome 25: The drawn ticket is the one which is marked with '25'
So there are 25 outcomes. We need not write them all. In fact, once we understand the basics, we need to write only this:
Step 1: There are 25 possible outcomes.
Step 2: Analyse each outcome. According to the question, the card should be a multiple of 3 OR 7. So if the card drawn is either a multiple of 3 OR a multiple of 7, we can write 'favourable' towards that outcome. And we will have an event. Let us see on which all out comes, we can write 'favourable':
Ticket marked with '1' → 1 is not a multiple of 3 or 7 → Not favourable
Ticket marked with '2' → 2 is not a multiple of 3 or 7 → Not favourable
Ticket marked with '3' → 3 is a multiple 3 → Favourable
Ticket marked with '4' → 4 is not a multiple of 3 or 7 → Not favourable
Ticket marked with '5' → 5 is not a multiple of 3 or 7 → Not favourable
Ticket marked with '6' → 6 is a multiple 3 → Favourable
Ticket marked with '7' → 7 is a multiple 7 → Favourable
- - -
- - -
Ticket marked with '25' → 25 is not a multiple of 3 or 7 → Not favourable
In this step 2 also, it is not necessary to write the 25 lines. We need to write only this:
Step 2: Analysing each outcome, we find that, 3, 6, 7, 9, 12, 14, 15, 18, 21, and 24 satisfy the required conditions. So there are a total of 10 favourable outcomes.
Thus the required probability = 10⁄25 = 2⁄5 = 40%. So there is a 40% probability that we will get a ticket which is either a multiple of 3 or a multiple of 7.
Solved example 1.11
Two unbiased coins are tossed simultaneously. Find the probability of getting (i) Two heads (ii) One tail (iii) One head (iv) at most one head (v) at least one head (vi) no head
Solution:
Step 1: Write the outcomes:
When two coins are tossed simultaneously, the possible outcomes can be written in the form of a table:
The above 4 are the only possible outcomes. This table can be used for all 6 cases of the question.
(i) Two heads: Analysing each outcome, we find that only one outcome (Outcome 1) satisfies this condition. So the probability of getting 2 heads = 1⁄4 = 25%
(ii) One tail: Two outcomes (Outcome 2 and 4) satisfies this condition. So the probability = 2⁄4 = 1⁄2 = 50%
(iii) One head: Two outcomes (Outcome 2 and 4) satisfies this condition. So the probability = 2⁄4 = 1⁄2 = 50%
(iv) At most one head: This condition means that when two coins are tossed simultaneously, there must be only one head. There must not be more than one head.
Two outcomes (Outcome 2 and 4) satisfies this condition. So the probability = 2⁄4 = 1⁄2 = 50%
(v) At least one head: This condition means that, when two coins are tossed simultaneously, there must be atleast one head. So even if there are two heads, the condition will be satisfied. Three outcomes (outcomes 1, 2 and 4) satisfies this condition. So the probability = 3⁄4 = 75%
(vi) No head: This condition means that when two coins are tossed simultaneously, there must not be any head. Only one outcome (Outcome 3) satisfies this condition. So the probability = 1⁄4 = 25%
Solved example 1.12
Three unbiased coins are tossed simultaneously. Find the probability of getting (i) All heads (ii) two heads (iii) One head (iv) at least two heads
Solution:
Step 1: Write the outcomes:
When three coins are tossed simultaneously, the possible outcomes can be written in the form of a table:
The above 8 are the only possible outcomes. This table can be used for all 4 cases of the question.
(i) All heads: Analysing each outcome, we find that only one outcome (Outcome 1) satisfies this condition. So we can write: The probability of getting all heads = 1⁄8 = 12.5%
(ii) Two heads: Three outcomes (outcomes 2, 6 and 8) satisfies this condition. So the probability = 3⁄8 = 37.5%
(iii) One head: Three outcomes (Outcome 3, 5 and 7) satisfies this condition. So the probability = 3⁄8 = 37.5%
(iv) At least two heads: This condition means that when three coins are tossed simultaneously, there must be atleast two heads. So even if there are 3 heads, the condition will be satisfied.
Four outcomes (Outcome 1, 2, 6 and 8) satisfies this condition. So the probability = 4⁄8 = 1⁄2 = 50%
In the next section we will see some more solved examples
Solved example 1.10
Tickets marked with numbers from 1 to 25 are put in a box and mixed well. Then one ticket is drawn from the box without looking. What is the probability that the drawn ticket is a multiple of 3 or 7?
Solution:
Step 1: Write the outcomes:
Outcome 1: The drawn ticket is the one which is marked with '1'
Outcome 2: The drawn ticket is the one which is marked with '2'
Outcome 3: The drawn ticket is the one which is marked with '3'
- - -
- - -
Outcome 25: The drawn ticket is the one which is marked with '25'
So there are 25 outcomes. We need not write them all. In fact, once we understand the basics, we need to write only this:
Step 1: There are 25 possible outcomes.
Step 2: Analyse each outcome. According to the question, the card should be a multiple of 3 OR 7. So if the card drawn is either a multiple of 3 OR a multiple of 7, we can write 'favourable' towards that outcome. And we will have an event. Let us see on which all out comes, we can write 'favourable':
Ticket marked with '1' → 1 is not a multiple of 3 or 7 → Not favourable
Ticket marked with '2' → 2 is not a multiple of 3 or 7 → Not favourable
Ticket marked with '3' → 3 is a multiple 3 → Favourable
Ticket marked with '4' → 4 is not a multiple of 3 or 7 → Not favourable
Ticket marked with '5' → 5 is not a multiple of 3 or 7 → Not favourable
Ticket marked with '6' → 6 is a multiple 3 → Favourable
Ticket marked with '7' → 7 is a multiple 7 → Favourable
- - -
- - -
Ticket marked with '25' → 25 is not a multiple of 3 or 7 → Not favourable
In this step 2 also, it is not necessary to write the 25 lines. We need to write only this:
Step 2: Analysing each outcome, we find that, 3, 6, 7, 9, 12, 14, 15, 18, 21, and 24 satisfy the required conditions. So there are a total of 10 favourable outcomes.
Thus the required probability = 10⁄25 = 2⁄5 = 40%. So there is a 40% probability that we will get a ticket which is either a multiple of 3 or a multiple of 7.
Solved example 1.11
Two unbiased coins are tossed simultaneously. Find the probability of getting (i) Two heads (ii) One tail (iii) One head (iv) at most one head (v) at least one head (vi) no head
Solution:
Step 1: Write the outcomes:
When two coins are tossed simultaneously, the possible outcomes can be written in the form of a table:
| Outcome | First coin | Second coin |
|---|---|---|
| Outcome 1 | Heads | Heads |
| Outcome 2 | Heads | Tails |
| Outcome 3 | Tails | Tails |
| Outcome 4 | Tails | Heads |
The above 4 are the only possible outcomes. This table can be used for all 6 cases of the question.
(i) Two heads: Analysing each outcome, we find that only one outcome (Outcome 1) satisfies this condition. So the probability of getting 2 heads = 1⁄4 = 25%
(ii) One tail: Two outcomes (Outcome 2 and 4) satisfies this condition. So the probability = 2⁄4 = 1⁄2 = 50%
(iii) One head: Two outcomes (Outcome 2 and 4) satisfies this condition. So the probability = 2⁄4 = 1⁄2 = 50%
(iv) At most one head: This condition means that when two coins are tossed simultaneously, there must be only one head. There must not be more than one head.
Two outcomes (Outcome 2 and 4) satisfies this condition. So the probability = 2⁄4 = 1⁄2 = 50%
(v) At least one head: This condition means that, when two coins are tossed simultaneously, there must be atleast one head. So even if there are two heads, the condition will be satisfied. Three outcomes (outcomes 1, 2 and 4) satisfies this condition. So the probability = 3⁄4 = 75%
(vi) No head: This condition means that when two coins are tossed simultaneously, there must not be any head. Only one outcome (Outcome 3) satisfies this condition. So the probability = 1⁄4 = 25%
Solved example 1.12
Three unbiased coins are tossed simultaneously. Find the probability of getting (i) All heads (ii) two heads (iii) One head (iv) at least two heads
Solution:
Step 1: Write the outcomes:
When three coins are tossed simultaneously, the possible outcomes can be written in the form of a table:
The above 8 are the only possible outcomes. This table can be used for all 4 cases of the question.
(i) All heads: Analysing each outcome, we find that only one outcome (Outcome 1) satisfies this condition. So we can write: The probability of getting all heads = 1⁄8 = 12.5%
(ii) Two heads: Three outcomes (outcomes 2, 6 and 8) satisfies this condition. So the probability = 3⁄8 = 37.5%
(iii) One head: Three outcomes (Outcome 3, 5 and 7) satisfies this condition. So the probability = 3⁄8 = 37.5%
(iv) At least two heads: This condition means that when three coins are tossed simultaneously, there must be atleast two heads. So even if there are 3 heads, the condition will be satisfied.
Four outcomes (Outcome 1, 2, 6 and 8) satisfies this condition. So the probability = 4⁄8 = 1⁄2 = 50%
In the next section we will see some more solved examples
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