A fair coin is tossed three times and the outcomes are noted. What is the probability of getting exactly two heads ?<\/p>\n
[amp_mcq option1=”\\(\\frac{2}{3}\\)” option2=”\\(\\frac{1}{2}\\)” option3=”\\(\\frac{5}{8}\\)” option4=”\\(\\frac{3}{8}\\)” correct=”option4″]<\/p>\n
We are looking for the probability of getting exactly two heads.
\nThe outcomes with exactly two heads are:
\nHHT
\nHTH
\nTHH
\nThere are 3 favourable outcomes.<\/p>\n
The probability of an event is defined as (Number of favourable outcomes) \/ (Total number of possible outcomes). A fair coin is tossed three times and the outcomes are noted. What is the probability of getting exactly two heads ? [amp_mcq option1=”\\(\\frac{2}{3}\\)” option2=”\\(\\frac{1}{2}\\)” option3=”\\(\\frac{5}{8}\\)” option4=”\\(\\frac{3}{8}\\)” correct=”option4″] This question was previously asked in UPSC CAPF – 2024 Download PDFAttempt Online The correct option is D. A fair coin is tossed three times. Each toss … <\/p>\n
\nProbability of getting exactly two heads = (Number of outcomes with exactly two heads) \/ (Total number of outcomes)
\nProbability = 3 \/ 8.
\n<\/section>\n
\nIn this case, n=3 (three tosses), k=2 (exactly two heads), and p=0.5 (probability of getting a head with a fair coin).
\nP(X=2) = C(3, 2) * (0.5)\u00b2 * (1-0.5)\u00b3\u207b\u00b2
\nP(X=2) = C(3, 2) * (0.5)\u00b2 * (0.5)\u00b9
\nP(X=2) = C(3, 2) * (0.5)\u00b3
\nC(3, 2) = 3! \/ (2! * (3-2)!) = 3! \/ (2! * 1!) = (3 * 2 * 1) \/ ((2 * 1) * 1) = 3.
\n(0.5)\u00b3 = 0.5 * 0.5 * 0.5 = 0.125 = 1\/8.
\nP(X=2) = 3 * (1\/8) = 3\/8.
\nThis confirms the result obtained by listing the outcomes.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"