Solveeit Logo
Login

Question

Question: Find the probability of getting \(2\)or\(3\) tails when a coin is tossed four times. \[ (a)\,\...

Find the probability of getting 22or33 tails when a coin is tossed four times.

(a)   716                 (b)  916                (c)  58                  (d)  516  (a)\,\,\,\dfrac{7}{{16}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\\ (b)\,\,\dfrac{9}{{16}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\\ (c)\,\,\dfrac{5}{8}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\\ (d)\,\,\dfrac{5}{{16}} \\\
Explanation

Solution

Use binomial distribution and taking n=4,     r=2    or  r=3  n = 4,\,\,\,\,\,r = 2\,\,\,\,or\,\,r = 3\,\,to get the result. Probability of an event is:
P(E)=favourable outcomestotal  number  of  outcomesP\left( E \right) = \dfrac{{favourable{\text{ }}outcomes}}{{total\,\,number\,\,of\,\,outcomes}}.

Complete step by step answer:
(1) When a coin is tossed, probability of getting a head is=12  = \dfrac{1}{2}\,
Probability of getting a tail when a coin is tossed=12 = \dfrac{1}{2}
∴  p=12,  q=12\therefore \,\,p = \dfrac{1}{2},\,\,q = \dfrac{1}{2}
(2) Coin is tossed44times
∴  n=4\therefore \,\,n = 4
(3) Probability of getting 22or33tails,
r=3  or   r=2r = 3\,\,or\,\,\,r = 2
(4) Using formula of binomial distributionnCr (p)r (q)n−r ^n{C_r}\,{(p)^r}\,{(q)^{n - r}}\,
Here, n is the total number of ways.
r is required.
p is success in one case
q is failure in one case
(5) Here, n=4,  r=2  or  r=3,  p=12,  q=12n = 4,\,\,r = 2\,\,or\,\,r = 3,\,\,p = \dfrac{1}{2},\,\,q = \dfrac{1}{2}
(6) Using values in formula mentioned in step (4), we get
4C2(12)2(12)2+ 4C3(12)3(12)1^4{C_2}{\left( {\dfrac{1}{2}} \right)^2}{\left( {\dfrac{1}{2}} \right)^2} + {\,^4}{C_3}{\left( {\dfrac{1}{2}} \right)^3}{\left( {\dfrac{1}{2}} \right)^1}
=4!2! 2!×14×14+4!3! 1!×18×12= \dfrac{{4!}}{{2!\,2!}} \times \dfrac{1}{4} \times \dfrac{1}{4} + \dfrac{{4!}}{{3!\,1!}} \times \dfrac{1}{8} \times \dfrac{1}{2}
=4×3×2×12×1×2×1×14×14+4×3!3!×1×18×12= \dfrac{{4 \times 3 \times 2 \times 1}}{{2 \times 1 \times 2 \times 1}} \times \dfrac{1}{4} \times \dfrac{1}{4} + \dfrac{{4 \times 3!}}{{3! \times 1}} \times \dfrac{1}{8} \times \dfrac{1}{2}
=4×32×116+4×18×12= \dfrac{{4 \times 3}}{2} \times \dfrac{1}{{16}} + 4 \times \dfrac{1}{8} \times \dfrac{1}{2}
=38+14= \dfrac{3}{8} + \dfrac{1}{4}
=3+28= \dfrac{{3 + 2}}{8}
=58= \dfrac{5}{8}
There probability of getting two or three tails if the coin is tossed 44 times is 58\dfrac{5}{8}
Hence, the correct option is (c)  (58)(c)\,\,\left( {\dfrac{5}{8}} \right)

Additional Information: Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one. Probability has been introduced in Maths to predict how likely events are to happen.

Note: When an object is tossed n times, we use binomial distribution in probability.