A coin is tossed twice. If events A and B are | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(d) Total number of ways $ = (HH,\,HT,\,TH,\,TT)$

$P$ (head on first toss) $ = \frac{2}{4} = \frac{1}{2} = P(A)$

$P$ (head on second toss) $ = \

Vedclass · Accepted Answer

$\frac{3}{4}$

Vedclass · Answer

(d) Total number of ways $ = (HH,\,HT,\,TH,\,TT)$

$P$ (head on first toss) $ = \frac{2}{4} = \frac{1}{2} = P(A)$

$P$ (head on second toss) $ = \frac{2}{4} = \frac{1}{2} = P(B)$

$P$ (head on both toss) $ = \frac{1}{4} = P(A \cap B)$

Hence required probability is,

$P(A \cup B) = P(A) + P(B) - P(A \cap B) $

$= \frac{1}{2} + \frac{1}{2} - \frac{1}{4} = \frac{3}{4}$ .

A coin is tossed twice. If events $A$ and $B$ are defined as :$A =$ head on first toss, $B = $ head on second toss. Then the probability of $A \cup B = $

Similar Questions

In a class of $60$ students, $30$ opted for $NCC$ , $32$ opted for $NSS$ and $24$ opted for both $NCC$ and $NSS$. If one of these students is selected at random, find the probability that The student has opted $NSS$ but not $NCC$.

If $P(A)=\frac{3}{5}$ and $P(B)=\frac{1}{5},$ find $P(A \cap B)$ if $A$ and $B$ are independent events

A party of $23$ persons take their seats at a round table. The odds against two persons sitting together are

Let $A$ and $B $ be two events such that $P\left( {\overline {A \cup B} } \right) = \frac{1}{6}\;,P\left( {A \cap B} \right) = \frac{1}{4}$ and $P\left( {\bar A} \right) = \frac{1}{4}$ where $\bar A$ stands for the complement of the event $A$. Then the events $A$ and$B$ are

If $P\,(A) = 0.4,\,\,P\,(B) = x,\,\,P\,(A \cup B) = 0.7$ and the events $A$ and $B$ are mutually exclusive, then $x = $