• #Definition
• #Q1 — Two coins tossed
• #Q2 — Single dice
• #Q3 — Two dice, sum greater than 9
• #Q4 — Two dice, sum divisible by 5, 4
• #Cards
• #Q5 — Probability of drawing a King
• #Q6 — Probability of King or Queen
• #Q7 — One spade and one diamond
• #Q8 — Bag of coloured balls

Maths | PDF

/Probability — Wednesday, 21 April 2021, 8:40 AM/

\[ \text{Probability} \;=\; \frac{\text{No. of favourable outcomes}}{\text{Total outcomes}} \]

(1) Qs: 2 coins are tossed simultaneously, find the probability of getting -

(i) Head on both coins.

(ii) None head on both coins.

(iii) 1 Head and 1 Tail on both coins.

(iv) At least 1H on both coins.

(v) At most 1H on both coins.

Total no. of outcomes \((4)\)

\[ = \{HH,\,HT,\,TH,\,TT\} \]

(i) \(P(HH) = 1/4\)

(ii) \(P(TT) = 1/4\)

(iii) \(P(HT\ \text{or}\ TH) = 2/4 = 1/2\)

(iv) \(P(\text{at least one head})\)

\[ = 1 - P(\text{none}) \] \[ = 1-P(TT) \] \[ = 1-1/4 \] \[ = 3/4 \]

(v) \(P(\text{At most one head})\)

\[ = \frac{N\big((H,T)\,(T,H)\,(T,T)\big)}{4} \] \[ P = 3/4 \]

(2)    DICE:- \(1/2/3/4/5/6\)     \(0/2/4/6/8\)

One dice rolled then what is the probability -

(i) even no. comes in front face \(\Rightarrow \dfrac{3}{6} = 1/2\)   Ans

/[The next two sub-parts continue on a pasted photo of a whiteboard, not in the student's own hand.]/

(ii) No. should be divisible by 2 \(\Rightarrow \dfrac{3}{6} = 1/2\)   Ans

(iii) No. should be greater than 2 but lesser than 4.

\[ \Rightarrow \frac{1}{6}\ \text{Ans} \]

(3)    DICE:-

Ques: Two dice are thrown simultaneously find the probability of getting -

(i) Sum is greater than 9 on the face of both the dice?

Favourable outcomes

\[ = \big\{(6,4)\,(4,6),\,(6,5)\,(5,6),\,(6,6)\,(5,5)\big\} \] \[ P = \frac{6}{6\times6} = \frac{1}{6} \]

Total possible outcomes: \(6\times6\) — direct product of space of states.

(4)    DICE:-

Ques: Two dice are thrown simultaneously find the probability of getting -

(ii) Sum is divisible by 5.

(iii) Sum is divisible by 4 but not by 3.

(i,i) Sum \((6,1)\)   Case: \(2\leftrightarrow12\to1\), \(3\leftrightarrow11\to2\), \(4\leftrightarrow10\to3\), \(5\leftrightarrow9\to\ldots\) (running count of ways to make each total)

(ii) Sum \(= 5,\,10\)

\[ \longrightarrow \{(5,5)\,(4,6)\,(6,4)\} \] \[ \{(2,3)\,(3,2)\,(1,4)\,(4,1)\} \] \[ \text{Probability} = \frac{3+4}{6\times6} = \frac{7}{36} \]

(iii) Sum is divisible by 4 but not by 3

Required outcomes \(= 4,\,8\)

\[ \longrightarrow \big\{(4,4),\,(5,3)\,(3,5)\,(2,6)\,(6,2)\big\} \] \[ \big\{(1,3)\,(3,1)\,(2,2)\big\} \] \[ \text{Probability} = \frac{8}{36} = \frac{2}{9} \]

Total cards \(= 52\)   \((13+13+13+13)\)

\begin{tikzpicture}[scale=1.0]
  \foreach \i/\txt in {0/Heart,1/Diamonds,2/Club,3/Spade} {
    \draw (\i*1.7,0) rectangle (\i*1.7+1.5,1.0);
    \node at (\i*1.7+0.75,0.5) {\small \txt};
  }
  \node[above] at (3.4,1.15) {\small POKER CARD SYMBOL};
\end{tikzpicture}

\[ \text{Face cards } (12): \quad \left.\begin{aligned}\text{King}\\\text{Queen}\\\text{Jack}\end{aligned}\right\} \]

Ace, together with King/Queen/Jack, \(\to\) Honour Cards \((16)\)

(5) Ques: A card is drawn from a pack of 52 cards, find the probability that the card drawn is King?

Sol\(^n\):

\[ P = \frac{4\ kings}{52\ cards} = \frac{1}{13} \]

(ii) Find the probability of getting 2 kings when withdrawing 2 cards

\[ P = \frac{\text{fav. outcomes}}{\text{total outcomes}} = \frac{{}^{4}C_2}{{}^{52}C_2} \]

\(({}^{4}C_2 \leftarrow\) No of ways of selecting 2 kings from 4 kings; \({}^{52}C_2 \leftarrow\) No of ways of selecting 2 cards\()\)

\[ P = \frac{\dfrac{4!}{2!\,2!}}{\dfrac{52!}{50!\,2!}} = \frac{4\times3\times2}{52\times51} = \frac{1}{13\times17} \] \[ P = \frac{1}{221} \]

(6) A card is drawn from a pack of 52 cards. Find the probability that the card drawn is King or Queen?

\[ P = \frac{\text{fav. outcomes}}{\text{total outcomes}} = \frac{8}{52} = \frac{2}{13} \]

or

\[ P = \frac{\text{No of ways of getting King or Queens}}{\text{No of ways of selecting 1 card from 52}} = \frac{{}^{4}C_1+{}^{4}C_1}{{}^{52}C_1} \] \[ = \frac{4+4}{52} = \frac{2}{13} \]

(7) Two cards are drawn from a pack of 52 Cards. Find the probability that one card is from Spade and one from Diamond.

fav. outcomes \(= 13\times13 = 169\)

\[ P = \frac{\text{fav. outcomes}}{\text{total outcomes}} = \frac{13\times13}{{}^{52}C_2} = \frac{{}^{13}C_1\times{}^{13}C_1}{{}^{52}C_2} \] \[ P = \frac{13\times13\times2}{52\times51} = \frac{13}{102} \]

(8) Ques: A bag contains 3 Red balls, 4 Black Balls and 5 green balls. 3 Balls are drawn randomly. Find the probability of getting -

\(3R+4B+5G = 12\)   (3 balls are drawn randomly)

(i) All the balls are different colours.

\[ P = \frac{\text{fav outcomes}}{\text{total outcomes}} = \frac{N(\text{All balls of diff colors})}{N(3\ \text{Balls drawn})} \]

\begin{tikzpicture}[scale=1.0]
  \foreach \i/\txt in {0/12,1/11,2/10} {
    \draw (\i,0) rectangle (\i+1,0.9);
    \node at (\i+0.5,0.45) {$\txt$};
  }
\end{tikzpicture}

\[ \frac{12\times11\times10}{3!} = {}^{12}C_3 \] \[ = \frac{N(1\ \text{Red and}\ 1\ \text{Black and}\ 1\ \text{Green})}{{}^{12}C_3} \]

/[As written, this line reads "1 Red and 1 Blue and 1 Green" — a slip for Black, since the bag has no blue balls; parts (ii)–(vi) below correctly call it Black.]/

\[ = \frac{{}^{3}C_1\times{}^{4}C_1\times{}^{5}C_1}{{}^{12}C_3} = \frac{3\times4\times5\times3\times2}{12\times11\times10} \qquad (3\times2 = \text{No. of permutation of 3 objects}) \] \[ \left(P = \frac{3}{11}\right)\ \checkmark \]

(ii) \(P(\text{All the balls are of same colour})\)

\[ = \frac{N(RRR\ \text{or}\ GGG\ \text{or}\ BBB)}{\text{No of total ways of selecting 3}} = \frac{{}^{3}C_3+{}^{5}C_3+{}^{4}C_3}{{}^{12}C_3} \] \[ = \frac{1+\dfrac{5\times4}{2}+4}{\dfrac{12\times11\times10}{3!}} = \frac{15\times3\times2}{12\times11\times10} = \frac{3}{44} \]

(iii) \(P(\text{None Balls are of Black colour})\)

\[ = \frac{N\big((2G\,\&\,1R)\ \text{or}\ (2R\,\&\,1G)\ \text{or}\ 3G\ \text{or}\ 3R\big)}{\text{Total no. of outcomes.}} \] \[ = \frac{{}^{5}C_2\times{}^{3}C_1 + {}^{3}C_2\times{}^{5}C_1 + {}^{5}C_3 + {}^{3}C_3}{{}^{12}C_3} \] \[ = \frac{\frac{5\times4}{2}\times3 + 3\times5 + \frac{5\times4}{2} + 1}{\dfrac{12\times11\times10}{3\times2}} = \frac{(30+15+10+1)\times6}{12\times11\times10} \] \[ P = \frac{56\times6}{12\times11\times10\times\cancel{5}} = \frac{14}{55} \]

(OR)   \(P(\text{None Balls are Black}) = \dfrac{{}^{8}C_3}{{}^{12}C_3} = \dfrac{14}{55}\)

(iv) \(P(\text{All the Balls should not be black})\)

\[ = 1-P(\text{All balls Black}) \] \[ = 1 - \frac{BBB}{{}^{12}C_3} = 1-\frac{{}^{4}C_3}{{}^{12}C_3} \] \[ P = 1-\frac{4\times3\times2}{12\times11\times10} = 1-\frac{1}{55} = \frac{54}{55} \]

(v) \(P(\text{At least one ball is green}) = 1-P(\text{No green balls})\)

\[ = 1-\frac{{}^{7}C_3}{{}^{12}C_3} \] \[ P = 1-\frac{7\times6\times5\times3\times2}{3\times2\times12\times11\times10} \] \[ P = 1-\frac{7}{44} = \frac{37}{44} \]

(vi) \(P(\text{At most one ball is green})\)

\[ = 1-P(2\ \text{balls green})-P(3\ \text{balls green}) \] \[ = 1-\frac{N\big((2G\,\&\,1B)+(2G\,\&\,1R)\big)}{{}^{12}C_3} - \frac{{}^{5}C_3}{{}^{12}C_3} \] \[ = 1-\frac{{}^{5}C_2\times{}^{4}C_1 + {}^{5}C_2\times{}^{3}C_1 + {}^{5}C_3}{{}^{12}C_3} \] \[ = 1-\frac{\frac{5\times4}{2}\times4 + \frac{5\times4}{2}\times3 + \frac{5\times4}{2}}{\dfrac{12\times11\times10}{6}} \] \[ = 1-\frac{(40+30+10)\times6}{12\times11\times10} = 1-\frac{80\times6}{12\times11\times10} \] \[ \left(P = 1-\frac{4}{11} = \frac{7}{11}\right)\ \checkmark \]

(OR)   \(P(\text{At most one ball is green}) = \dfrac{N\big((1G\,\&\,2\,\text{other})+(0G\,\&\,3\,\text{other})\big)}{\text{Total outcomes}}\)

\[ = \frac{{}^{5}C_1\times{}^{7}C_2 + {}^{5}C_0\times{}^{7}C_3}{{}^{12}C_3} \] \[ = \frac{5\times7\times6/2 + 1\times7\times6\times5/6}{{}^{12}C_3} \] \[ \left(P = \frac{140}{220} = \frac{7}{11}\right)\ \checkmark \]