The probability that at least one of the two events A and B occurs is 0.6. If A and B occur simultaneously wit

NCERT Exemplar · Q3

A

Assam AhsecShort· 3mImportance★★★★★

53% · 88/165 Questions

✓ Free question

Concept understanding — Probability Complement Rule

The Probability Complement Rule

Every event $A$ splits the sample space in two: outcomes where $A$ happens, and outcomes where it does not. The second group is the complement of $A$ , written $A'$ (also $A^c$ or $\bar{A}$ ). Because exactly one of the two must occur, their probabilities together fill the whole space:

P(A) + P(A') = 1 \quad\Longrightarrow\quad P(A') = 1 - P(A).

That is the complement rule: the probability that $A$ does not happen is $1$ minus the probability that it does.

Why It Holds

$A$ and $A'$ are mutually exclusive (no outcome lies in both) and exhaustive (together they are the entire sample space $S$ , with $P(S) = 1$ ). So $P(A) + P(A') = P(S) = 1$ , and rearranging gives the rule.

A Simple Example

For a fair die, $P(\text{six}) = \tfrac{1}{6}$ , so $P(\text{not six}) = 1 - \tfrac{1}{6} = \tfrac{5}{6}$ .

Why It Is So Useful: the "At Least One" Trick

Counting "at least one" directly often means adding many separate cases, while its complement, "none," is a single easy case. For instance, the probability of at least one head in three tosses of a fair coin:

P(\text{at least one head}) = 1 - P(\text{no heads}) = 1 - \left(\tfrac{1}{2}\right)^3 = 1 - \tfrac{1}{8} = \tfrac{7}{8}.

Computing "no heads" once is far quicker than summing the one-head, two-head and three-head cases separately.

Tip

Whenever a question asks for the probability of "at least one," pause and try $1 - P(\text{none})$ first — it usually turns a long sum into a one-line calculation.

Note

The rule combines with others too: $P(A' \cap B') = 1 - P(A \cup B)$ , which is how De Morgan's laws appear in probability.

The complement rule and its "at least one" shortcut are staples of the NCERT Class 12 Probability chapter, tested constantly in CBSE boards, JEE Main and state CETs wherever a question asks for P(at least one). Students searching "probability of at least one event formula" will find this trick turns some of the hardest-looking probability questions into one-line calculations.

The key idea is to use the complement rule: $P(A') + P(B') = 2 - [P(A) + P(B)]$ . From the given data, $P(A \cup B) = 0.6$ and $P(A \cap B) = 0.3$ , so $P(A) + P(B) = P(A \cup B) + P(A \cap B) = 0.9$ . Thus $P(A') + P(B') = 2 - 0.9 = 1.1$ .

The problem asks for $P(A') + P(B')$ , the sum of the probabilities of the complements of two events. A direct approach would require knowing $P(A)$ and $P(B)$ individually, but we are not given those. Instead, we are given two pieces of information:

$P(A \cup B) = 0.6$ — the probability that at least one occurs.
$P(A \cap B) = 0.3$ — the probability that both occur simultaneously.

The complement rule tells us that $P(A') = 1 - P(A)$ and $P(B') = 1 - P(B)$ . So:

P(A') + P(B') = (1 - P(A)) + (1 - P(B)) = 2 - [P(A) + P(B)].

The problem reduces to finding $P(A) + P(B)$ from the given union and intersection. This is where the addition rule of probability comes in.

A

Rearranging:

P(A) + P(B) = P(A \cup B) + P(A \cap B).

Now substitute the given values:

$P(A \cup B) = 0.6$
$P(A \cap B) = 0.3$

So:

P(A) + P(B) = 0.6 + 0.3 = 0.9.

Therefore:

P(A') + P(B') = 2 - 0.9 = 1.1.

Watch out

A common mistake is to think $P(A') + P(B') = 1 - P(A \cup B)$ or something similar. But complements don't combine that way — you must go through $P(A) + P(B)$ .

Tip

Notice that we never needed $P(A)$ or $P(B)$ individually. The sum $P(A) + P(B)$ was enough. This is a neat trick: whenever you see $P(A') + P(B')$ , think $2 - [P(A) + P(B)]$ , and use the addition rule to get the sum.

✓Final answer

P(A') + P(B')

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AHSEC Higher Secondary (HS) Final Examination 2019Set ANNUAL6 marks
$3\times3$
›Reveal solutionSolution

Main: it's easier to find $P(\text{product NOT divisible by }3)=P(\text{both chosen numbers not divisible by }3)$ and subtract from $1$ . OR: for one row to form a 3-digit-style number divisible by $15$ requires divisibility by both $5$ and $3$ ; count favourable digit combinations, then cube the single-row probability since the $9$ entries are chosen independently.

Main question. Two numbers are selected (without replacement) from the first $90$ natural numbers $\{1,\dots,90\}$ . Find $P(\text{product divisible by }3)$ .

Total ways to choose 2 numbers: $\binom{90}{2} = \dfrac{90\times89}{2} = 4005$ .

Among $1$ to $90$ , the multiples of $3$ are $3,6,\dots,90$ — that's $30$ numbers; the remaining $60$ are not multiples of $3$ .

The product of the two chosen numbers is divisible by $3$ unless neither number is a multiple of $3$ (i.e. the product is divisible by 3 iff at least one factor is).

$P(\text{neither divisible by }3) = \frac{\binom{60}{2}}{\binom{90}{2}} = \frac{1770}{4005}.$

$P(\text{product divisible by }3) = 1 - \frac{1770}{4005} = \frac{2235}{4005} = \frac{149}{267} \approx 0.558$

(dividing numerator and denominator by $15$ : $2235/15=149$ , $4005/15=267$ ; $149$ is prime and doesn't divide $267=3\times89$ , so this is fully reduced).

OR question. Entries $a_{ij}$ of a $3\times3$ matrix are chosen at random from the digits $0,1,\dots,9$ with replacement, and each row's three digits (read left to right) are treated as a 3-digit number. Find the probability that every row's number is divisible by $15$ .

Since all $9$ entries are chosen independently, and the three rows are structurally identical, the probability that all three rows individually satisfy the divisibility condition is $\left[P(\text{one row divisible by }15)\right]^3$ .

One row: digits $d_1d_2d_3$ , each uniform over $\{0,\dots,9\}$ independently (so 1000 equally likely 3-digit strings from $000$ to $999$ ). Divisibility by $15=3\times5$ requires divisibility by both $3$ and $5$ .

The multiples of $15$ in the range $000$ to $999$ are $0,15,30,\dots,990$ — that's $\left\lfloor\dfrac{990}{15}\right\rfloor+1 = 66+1=67$ values.

So $P(\text{one row divisible by }15) = \dfrac{67}{1000}$ .

Since the three rows are independent:

$P(\text{all three rows divisible by }15) = \left(\frac{67}{1000}\right)^3 = \frac{300763}{1000000000} \approx 0.0003.$

(Note: the source stem's phrasing "each element $a_{ij}$ is a three digit number" is unusual — a single digit $0$ – $9$ can't itself be a three-digit number — so this solution interprets the intended meaning as: each row's 3 single-digit entries, read together, form a 3-digit string, and we want each such string divisible by 15. The digit-counting method above is presented explicitly so the approach is checkable regardless of the exact original phrasing.)

✓Final answer
$P(\text{product divisible by }3) = \dfrac{149}{267}$

←Previous questionQ2 — Refer to Exercise 1 above. If the die were fair, determine whether or …→Next questionQ4 — A bag contains 5 red marbles and 3 black marbles. Three marbles are dr…

Q.The probability that at least one of the two events $A$ and $B$ occurs is $0.6$ . If $A$ and $B$ occur simultaneously with probability $0.3$ , evaluate $P(A') + P(B')$ .

Concept understanding — Probability Complement Rule

The Probability Complement Rule

Why It Holds

A Simple Example

Why It Is So Useful: the "At Least One" Trick

Method: Relating Complement Sums to the Addition Rule

Steps

Common Mistakes

Q.The probability that at least one of the two events AAA and BBB occurs is 0.60.60.6. If AAA and BBB occur simultaneously with probability 0.30.30.3, evaluate P(A′)+P(B′)P(A') + P(B')P(A′)+P(B′).

🎓Unlock everything free for 14 days

Q.The probability that at least one of the two events $A$ and $B$ occurs is $0.6$ . If $A$ and $B$ occur simultaneously with probability $0.3$ , evaluate $P(A') + P(B')$ .

Unlock everything free for 14 days