Here we will learn how to find the probability of tossing two coins. Let us take the experiment of tossing two coins simultaneously: When we toss two coins simultaneously then the possible of outcomes are: (two heads) or (one head and one tail) or (two tails) i.e., in short (H, H) or (H, T) or (T, T) respectively; where H is denoted for head and T is denoted for tail. Therefore, total numbers of outcome are 22 = 4The above explanation will help us to solve the problems on finding the probability of tossing two coins. Worked-out problems on probability involving tossing or flipping two coins: 1. Two different coins are tossed randomly. Find the probability of: (i) getting two heads (ii) getting two tails (iii) getting one tail (iv) getting no head (v) getting no tail (vi) getting at least 1 head (vii) getting at least 1 tail (viii) getting atmost 1 tail (ix) getting 1 head and 1 tail Solution: When two different coins are tossed randomly, the sample space is given by S = {HH, HT, TH, TT} Therefore, n(S) = 4. (i) getting two heads: Let E1 = event of getting 2 heads. Then,E1 = {HH} and, therefore, n(E1) = 1. Therefore, P(getting 2 heads) = P(E1) = n(E1)/n(S) = 1/4. (ii) getting two tails: Let E2 = event of getting 2 tails. Then,E2 = {TT} and, therefore, n(E2) = 1. Therefore, P(getting 2 tails) = P(E2) = n(E2)/n(S) = 1/4. (iii) getting one tail: Let E3 = event of getting 1 tail. Then,E3 = {TH, HT} and, therefore, n(E3) = 2. Therefore, P(getting 1 tail) = P(E3) = n(E3)/n(S) = 2/4 = 1/2 (iv) getting no head: Let E4 = event of getting no head. Then,E4 = {TT} and, therefore, n(E4) = 1. Therefore, P(getting no head) = P(E4) = n(E4)/n(S) = ¼. (v) getting no tail: Let E5 = event of getting no tail. Then,E5 = {HH} and, therefore, n(E5) = 1. Therefore, P(getting no tail) = P(E5) = n(E5)/n(S) = ¼. (vi) getting at least 1 head: Let E6 = event of getting at least 1 head. Then,E6 = {HT, TH, HH} and, therefore, n(E6) = 3. Therefore, P(getting at least 1 head) = P(E6) = n(E6)/n(S) = ¾. (vii) getting at least 1 tail: Let E7 = event of getting at least 1 tail. Then,E7 = {TH, HT, TT} and, therefore, n(E7) = 3. Therefore, P(getting at least 1 tail) = P(E2) = n(E2)/n(S) = ¾. (viii) getting atmost 1 tail: Let E8 = event of getting atmost 1 tail. Then,E8 = {TH, HT, HH} and, therefore, n(E8) = 3. Therefore, P(getting atmost 1 tail) = P(E8) = n(E8)/n(S) = ¾. (ix) getting 1 head and 1 tail: Let E9 = event of getting 1 head and 1 tail. Then,E9 = {HT, TH } and, therefore, n(E9) = 2. Therefore, P(getting 1 head and 1 tail) = P(E9) = n(E9)/n(S)= 2/4 = 1/2. The solved examples involving probability of tossing two coins will help us to practice different questions provided in the sheets for flipping 2 coins. Probability Probability Random Experiments Experimental Probability Events in Probability Empirical Probability Coin Toss Probability Probability of Tossing Two Coins Probability of Tossing Three Coins Complimentary Events Mutually Exclusive Events Mutually Non-Exclusive Events Conditional Probability Theoretical Probability Odds and Probability Playing Cards Probability Probability and Playing Cards Probability for Rolling Two Dice Solved Probability Problems Probability for Rolling Three Dice 9th Grade Math From Probability of Tossing Two Coins to HOME PAGE
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1A number is selected at random from the first 50 positive integers. What is the probability that it is a Prime number? Answer: A Explanation: There are 15 Prime numbers from 1 to 50. $\therefore$ Required probability $ = \dfrac{{15}}{{100}} = \dfrac{3}{{20}}$ 2Four Fair coins are tossed together. Find the probability that tail appears on exactly one of the coins. A2/4 B3/4 C1/4 D2/5 Answer: C Explanation: Total number of cases when $n$ coins are tossed $=2^n$ The total number of cases when 4 coins are tossed $= 2^4\; = 16$ As Tail has to appear on exactly one of the coins, the favourable cases $= THHH, HTHH, HHTH, HHHT$. $\therefore$ Required probability $ = \dfrac{4}{{16}} = \dfrac{1}{4}$ Alternative Method: Number of way of getting exactly $r$ tails or heads on tossing $n$ coins = $\dfrac{{{}^n{C_r}}}{{{2^n}}}$ $\therefore$ $\dfrac{{{}^4{C_1}}}{{{2^4}}} = \dfrac{4}{{16}} = \dfrac{1}{4}$3 Four Fair coins are tossed together. Find the probability that tail appears on exactly two of the coins. A7/8 B3/4 C5/8 D3/8 Answer: D Explanation: Total number of cases when $n$ coins are tossed $=2^n$ The total number of cases $= 2^4\; = 16$ Tail has to appear on exactly 2 coins, i.e $TTHH$ in some order. Total ways of arranging two $H$'s and two $T$'s = $\dfrac{{4!}}{{2! \times 2!}}$ $= \dfrac{{4 \times 3 \times 2 \times 1}}{{2 \times 2}} = 6$ $\bigl($Formula:Number of ways of arranging $n$ items of which $p$ are alike, $q$ alike and so on $ = \dfrac{{n!}}{{p! \times q!}}\Biggr)$ $\therefore$Required probability $ = \dfrac{6}{{16}} = \dfrac{3}{8}$Alternative Method: Number of way of getting exactly $r$ tails or heads on tossing $n$ coins = $\dfrac{{{}^n{C_r}}}{{{2^n}}}$ $\therefore$ $\dfrac{{{}^4{C_2}}}{{{2^4}}} = \dfrac{6}{{16}} = \dfrac{3}{8}$4Eight unbiased coins are tossed simultaneously. What is the probability that at least 2 tails turn up? A233/256 B240/256 C222/256 D247/256 Answer: D Explanation: Number of way of getting exactly $r$ tails or heads on tossing $n$ coins = $\dfrac{{{}^n{C_r}}}{{{2^n}}}$ Probability of getting atleast 2 tails = 1 - (probability of getting no tails + probability of getting exactly one tail) $\therefore$ Probability that at least two tails turn up $ = 1 - \left( {\dfrac{{{}^8{C_0} + {}^8{C_1}}}{{{2^8}}}} \right)$ $=1 - \left( {\dfrac{{1 + 8}}{{256}}} \right) = 1 - \left( {\dfrac{9}{{256}}} \right)$ $={\dfrac{{247}}{{256}}}$ 5An unbiased dice is rolled. What is the probability that the number appearing on the dice is even? A1/3 B1/6 C1/2 D1/4 Answer: A Explanation: When a dice is rolled total outcomes $= 6$ Number of favourable outcomes $= 3(i.e., 2,4,6)$ $\therefore$Required probability $ = \dfrac{2}{6} = \dfrac{1}{3}$ 6When a fair cubical dice is rolled. What is the probability of getting a Prime number? A1/3 B1/4 C1/2 D2/3 Answer: C Explanation: On rolling a fair die, the total possibilities are 6. The number of ways of getting a Prime number $= 3(i.e., 2,3,5)$ $\therefore$The required probability = $ = \dfrac{3}{6} = \dfrac{1}{2}$ Page 2
7Two unbiased dice are thrown simultaneously. What is the probability that the sum of the numbers is 10? Answer: A Explanation: Total number of possible outcomes $=\; 6^2 = 36$ Number of outcomes that the sum of the numbers is 10 = 3 ($\because$ 5+5, 4+6, 6+4) $\therefore$ Required probability $ = \dfrac{3}{{36}} = \dfrac{1}{{12}}$ 8A bag contains 4 blue balls and 5 green balls. A ball is drawn at random. What is the probability that it is blue in colour? A2/9 B4/9 C5/9 D6/9 Answer: B Explanation: Total number number of ball $= 4 + 5 = 9$ Total number of ways of drawing one ball $ = {}^9{C_1} = 9$ Total number of ways of drawing one blue ball $ = {}^4{C_1} = 4$ $\therefore$ Required probability $ = \dfrac{4}{{9}}$ 9A bag contain 4 green marbles and 5 yellow marbles. Two marbles are drawn at random from the bag. What is the probability that both the balls are of same colour? A3/9 B5/9 C4/9 D6/9 Answer: C Explanation: Total number number of balls $= 4 + 5 = 9$ Number of ways in which the two marbles drawn are of same colour $ = {}^4{C_2} + {}^5{C_2}$$ = \dfrac{{4 \times 3}}{{2 \times 1}} + \dfrac{{5 \times 4}}{{2 \times 1}}$$ = 16$ Number of ways of drawing $2$ marbles out of $9$ marbles $ = {}^9{C_2} = \dfrac{{9 \times 8}}{{2 \times 1}} = 36$ $\therefore$Required probability $ = \dfrac{{16}}{{36}} = \dfrac{4}{9}$ 10If a card is drawn at random from a well shuffled pack of cards, find the probability of drawing a red card? A5/8 B2/8 C1/2 D4/6 Answer: C Explanation: Number of ways of selecting one card from a pack of cards $ = {}^{52}{C_1} = 52$ $\spadesuit$ = 13 Spades $\color{red}{\heartsuit}$ = 13 Hearts $\color{red}{\diamondsuit}$ = 13 Diamonds $\clubsuit$ = 13 Clubs Total red cards $= 13 + 13 = 26$ $\therefore$ Required probability $ = \dfrac{{26}}{{52}} = \dfrac{1}{2}$ 11If a card is drawn at random from a well shuffled pack of cards, find the probability of drawing an honour? A2/13 B3/13 C7/13 D4/13 Answer: D Explanation: $\spadesuit$ = 13 Spades; $\;\color{red}{\heartsuit}$ = 13 Hearts; $\;\color{red}{\diamondsuit}$ = 13 Diamonds; $\;\clubsuit$ = 13 Clubs Honour $=A,\;J,\;Q,\;K$ Each suite contains 4 honour cards mentioned above. Hence there are a total of 16 honour cards in a pack. $\therefore$ Required probability $ = \dfrac{{16}}{{52}} = \dfrac{4}{{13}}$ 12Two numbers are selected at random from the first 25 natural numbers. What is the probability that the sum of two numbers is 14? A1/50 B2/25 C2/50 D3/50 Answer: A Explanation: Number of outcomes that the sum of two numbers is 14 is $(1,13),(2,12)$,$(3,11),(4,10)$,$(5,9),(6,8)$ $= 6$ Number of ways of selecting two numbers out of 25 $ = {}^{25}{C_2} = \dfrac{{25 \times 24}}{{2 \times 1}} = 300$ $\therefore$ Required probability $ = \dfrac{6}{{300}} = \dfrac{1}{{50}}$ |
Eight unbiased coins are tossed simultaneously what is the probability that at least 2 tails turn up
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